# Ways of writing 100

There are eleven different ways of writing 100 in the form of mixed numbers using all the nine digits once and only once. Ten-of the ways have two figures in the integral part of the number, but the eleventh expression has only one figure there.

Can you find all the eleven expressions?

3 69258/714

81 5643/297

81 7524/396

82 3546/197

91 5742/638

91 5823/647

91 7524/836

94 1578/263

96 1428/357

96 1752/438

96 2148/537

I found these without writing a program.

The mixed fraction *i n*/*d* has integer portion *i* , numerator *n* , and denominator *d*. The integer portion *i* must have one or two digits, giving two possible partitions of the nine digits into *i*, *n* & *d*, with 9! = 362,880 ways to arrange the nine digits in each, giving a total initial search space of 725,760 arrangements for *i*, *n* & *d*:

- If
*i*is a single digit (*i1*), then*n*must be five digits (*n1*,*n2*,*n3*,*n4*,*n5*) and*d*three digits (*d1*,*d2*,*d3*) such that 91 <=*n*/*d*<= 99. The range of possible values for*n*/*d*are 12345/987 = 12 to 98765/123 = 802. The maximum value for*n*/*d*when the first digit of*n*is 1 is 19876/234 = 84, so the first digit of*n*can’t be 1. The minimum value for*n*/*d*when the first digit of*n*is 9 is 91234/876 = 104, so the first digit of*n*also can’t be 9. Since the first digit of*n*can’t be 1 or 9, there are only 7 * 8! = 282,240 possible permutations with a one-digit*i* - If
*i*is two digits (*i1*,*i2*), then*n*must be four digits (*n1*,*n2*,*n3*,*n4*) and*d*three digits (*d1*,*d2*,*d3*). The range of possible values for*n*/*d*are 1234/987 = 1 to 9876/123 = 80. Since*i*can’t be 99, the effective minimum is 2; since*i2*can’t be zero, the effective maximum value is 79. This give 21 <=*i*<= 98. Since*i1*can’t be 1, there are only 8 * 8! = 322,560 possible combinations with a two-digit*i*.

Now the search space is 282,240 + 322,560 = 604,800, which is a 17% reduction.

There are 9 * 8 * 7 = 504 combinations for the final digit of *i*, *n* and *d* (*dfi*, *dfn*, and *dfd* respectively) where each digit is a different value from 1 to 9. However, only 36 combinations satisfy the requirement that the final digit of (10 – *dfi*) * *dfd* = *dfn*. The possible combinations for (*dfi*, *dfn*, *dfd* ) are:

(1,2,8) (1,3,7) (1,4,6) (1,6,4) (1,7,3) (1,8,2)

(2,4,3) (2,4,8) (2,6,7) (2,8,1) (2,8,6)

(3,2,6) (3,4,2) (3,6,8) (3,7,1) (3,8,4) (3,9,7)

(4,2,7) (4,6,1) (4,8,3)

(6,2,3) (6,2,8) (6,4,1) (6,8,2) (6,8,7)

(7,2,4) (7,3,1) (7,4,8) (7,6,2) (7,8,6) (7,9,3)

(8,2,1) (8,2,6) (8,4,2) (8,4,7) (8,6,3)

It’s worth noting that no final digit can be 5 and neither *dfi* nor *dfd* may be 9 and *dfi* can’t be 1. This means for single-digit *i* that 92 <= *n*/*d* <= 99.

None of those combinations has both a 1 and a 9. 12 combinations have a 1, and 2 combinations contain a 9.

- For single-digit
*i*there are 5 * 5! = 600 ways to assign the remaining 6 digits for the 12 + 2 = 14 combinations that contain a 1 or 9; there are 4 * 5! = 480 ways to assign the remaining digits for the 36 – 14 = 22 combinations that don’t contain a 1 or 9. The search space for single-digit*i*is now 600 * 14 + 480 * 22 = 18,960. - For two-digit
*i*there are 6! = 720 ways to assign the remaining 6 digits for the 12 combinations that contain a 1; there are 5 * 5! = 600 ways to assign the remaining digits for the 36 – 12 = 24 combinations that don’t contain a 1. The search space for two-digit*i*is now 720 * 12 + 600 * 24 = 23,040.

Now the search space is 18,960 + 23,040 = 42,000, which is a 94% reduction from the initial search space.

From here, we can continue to prune the search space by cases that fix the assignment of one or more digits. Eventually, it comes down to checking the remaining possible permutations. For single-digit *i*, the most useful digit to assign first is the most-significant digit of *n*, which is already limited to 2..8. For two-digit *i*, the most useful digit to fix is the most-significant digit of *i*, which is already limited to 2..9. For those interested, the following discussion prunes down the possible values further. All division uses the floor value (rounding down) for maximum values or ceiling value (rounding up) for minimum values.

**Single-digit i, cases for n1:**

**2:** For 21346 <= *n* <= 29876, the range of values for *d* are 21346/99 = 216 to 29876/92 = 324. However, if *n1* = 2, then *d1* can’t be 2 and must be 3. This gives 314 <= *d* <= 318 (since *d2* can’t be 2 and *dfd* can’t be 9) making *d2* = 1, and making the minimum value for *n* = 92 * 314 = 28888. 1, 2 & 3 are assigned, and *n2* must be 8 or 9. There are only four combinations for *dfi*, *dfn*, and *dfd* that don’t have a 1, 2 or 3: (6,8,7), (7,4,8), (7,8,6) and (8,4,7). Each of those has an 8, making *n2* = 9. For each of the four combinations for *dfi*, *dfn*, and *dfd* there are 2! = 2 ways to assign *n3* and *n4*, giving 4 * 2 = 8 possibilities.

**3:** For 31246 <= *n* <= 39876, the range of values for *d* are 31246/99 = 316 to 39876/92 = 433. However, if *n1* = 3, then *d1* can’t be 3 and must be 4. This gives 412 <= *d* <= 428 (since *d2* can’t be 3 and *dfd* can’t be 9) making *d2* = 1 or 2, and making the minimum value for *n* = 92 * 412 = 37904. *n2* must be 7, 8 or 9. There are only 8 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 3 or 4 and also don’t have both 1 and 2: (2,6,7), (2,8,6), (6,2,8), (6,8,2), (6,8,7), (7,6,2), (7,8,6) and (8,2,6). Six of these have a 2, each of these has a 7 or 8, and two have both a 7 & 8. For *d2* = 1, there are 8 ways to assign *dfi*, *dfn*, and *dfd* if *n2* = 9, and 6 ways to assign *dfi*, *dfn*, and *dfd* if *n2* = 7 or 8, for a total of 2 * 6 + 8 = 20 possibilities for *d2* = 1. For *d2* = 2, *n2* = 9 since both combinations for *dfi*, *dfn*, and *dfd* that don’t have a 2–(6,8,7) and (7,8,6)–have both a 7 & 8. There are 20 + 2 = 22 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 3 and 2! = 2 ways to assign *n3* and *n4*, giving 22 * 2 = 44 possibilities.

**4:** For 41236 <= *n* <= 49876, the range of values for *d* are 41236/99 = 417 to 49876/92 = 542. However, if *n1* = 4, then *d1* can’t be 4 and must be 5. This gives 512 <= *d* <= 538 (since *d2* can’t be 4 and *dfd* can’t be 9) making *d2* = 1, 2 or 3 and making the minimum value for *n* = 92 * 512 = 47104. *n2* must be 7, 8 or 9. There are 22 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 4 or 5: (1,2,8), (1,3,7), (1,7,3), (1,8,2), (2,6,7), (2,8,1), (2,8,6), (3,2,6), (3,6,8), (3,7,1), (3,9,7), (6,2,3), (6,2,8), (6,8,2), (6,8,7), (7,3,1), (7,6,2), (7,8,6), (7,9,3), (8,2,1), (8,2,6) and (8,6,3). For *d2* = 1, there are {8, 6, 12} = 26 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. For *d2* = 2, there are {2, 6, 8} = 16 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. For *d2* = 3, there are {8, 2, 12} = 22 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. There are 26 + 16 + 22 = 64 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 4 and 2! = 2 ways to assign *n3* and *n4*, giving 64 * 2 = 128 possibilities.

**5:** For 51234 <= *n* <= 59876, the range of values for *d* are 51234/99 = 518 to 59876/92 = 650. However, if *n1* = 5, then *d1* can’t be 5 and must be 6. This gives 612 <= *d* <= 648 (since *dfd* can’t be 0 or 9) making *d2* = 1, 2, 3 or 4 and making the minimum value for *n* = 92 * 612 = 56304. *n2* can’t be 6 (since *d1* = 6), so the minimum value for *n* is 57123, making the minimum value for *d* = 57123/92 = 620. So now 621 <= *d* <= 648 (since *dfd* can’t be 0) making *d2* = 2, 3 or 4 and *n2* = 7, 8 or 9. There are 20 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 5 or 6: (1,2,8), (1,3,7), (1,7,3), (1,8,2), (2,4,3), (2,4,8), (2,8,1), (3,4,2), (3,7,1), (3,8,4), (3,9,7), (4,2,7), (4,8,3), (7,2,4), (7,3,1), (7,4,8), (7,9,3), (8,2,1), (8,4,2) and (8,4,7). For *d2* = 2, there are {2, 6, 8} = 16 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. For *d2* = 3, there are {6, 2, 10} = 18 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. For *d2* = 4, there are {4, 6, 8} = 18 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {7, 8, 9}. There are 16 + 18 + 18 = 52 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 5 and 2! = 2 ways to assign *n3* and *n4*, giving 52 * 2 = 104 possibilities.

**6:** For 61234 <= *n* <= 69874, the range of values for *d* are 61234/99 = 619 to 69874/92 = 759. However, if *n1* = 6, then *d1* can’t be 6 and must be 7. This gives 712 <= *d* <= 758 (since *dfd* can’t be 9) making *d2* = 1, 2, 3, 4 or 5 and making the minimum value for *n* = 92 * 712 = 65504. *n2* can’t be 6 or 7, so *n2* = 5, 8 or 9 and when *n2* = 5, *n3* = 8 or 9. There are 12 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 6 or 7: (1,2,8), (1,8,2), (2,4,3), (2,4,8), (2,8,1), (3,4,2), (3,8,4), (4,8,3), (8,2,1) and (8,4,2). For *d2* = 1, there are {6, 2, 6} = 14 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 8, 9}. For *d2* = 2, there are {2, 0, 2} = 4 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 8, 9}. For *d2* = 3, there are {6, 0, 6} = 12 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 8, 9}. For *d2* = 4, there are {4, 0, 4} = 8 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 8, 9}. For *d2* = 5, there are {2, 12} = 14 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {8, 9}. There are 14 + 4 + 12 + 8 + 14 = 52 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 6 and 2! = 2 ways to assign *n3* and *n4*, giving 52 * 2 = 104 possibilities. From this we can subtract the {4, 2, 6, 4} = 16 cases for *d2* = {1, 2, 3, 4} and *n2* = 5 where *n3* must be 9 (because 8 is assigned to *dfi*, *dfn*, or *dfd*). This gives 104 – 16 = 88 possibilities.

**7:** For 71234 <= *n* <= 79864, the range of values for *d* are 71234/99 = 720 to 79864/92 = 868. However, if *n1* = 7, then *d1* can’t be 7 and must be 8. However, if *n1* = 7 and *d1* = 8, then *i1* can’t be 7 or 8, and *n/d* can’t be 92 or 93. This gives a maximum value for *d* of 79864/94 = 849, and gives 812 <= *d* <= 836 (since *dfd* can’t be 7, 8 or 9) making *d2* = 1, 2 or 3 and making the minimum value for *n* = 94 * 812 = 76328. *n2* = 6 or 9. There are 8 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 7 or 8: (1,4,6), (1,6,4), (2,4,3), (3,2,6), (3,4,2), (4,6,1), (6,2,3) and (6,4,1). For *d2* = 1, there are {2, 4} = 6 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {6, 9}. For *d2* = 2, there are {0, 4} = 4 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {6, 9}. For *d2* = 3, there are {0, 3} = 3 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {6, 9}. There are 6 + 4 + 3 = 13 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 7 and 2! = 2 ways to assign *n3* and *n4*, giving 13 * 2 = 26 possibilities.

**8:** For 81234 <= *n* <= 89764, the range of values for *d* are 81234/99 = 821 to 89764/93 = 965 (if *n1* = 8, then *i1* can’t be 8, and *n/d* can’t be 92). However, if *n1* = 8, then *d1* can’t be 8 and must be 9. Since *n2* can’t be 8 or 9, the maximum value for *d* is the greater of 86543/93 = 930 (where *i1* = 7) or 87543/94 = 931 (where *i1* = 6). This gives 912 <= *d* <= 928 (because *dfd* can’t be 0 or 1) making *d2* = 1 or 2 and making the minimum value for *n* = 93 * 912 = 84816. However, since *n3* can’t be 8 or 9, the minimum value for *n* is actually 85123 and *n2* = 5, 6 or 7. There are 15 combinations for *dfi*, *dfn*, and *dfd* that don’t have an 8 or 9: (1,3,7), (1,4,6), (1,6,4), (2,4,3), (2,6,7), (3,2,6), (3,4,2), (3,7,1), (4,2,7), (4,6,1), (6,2,3), (6,4,1), (7,2,4), (7,3,1) and (7,6,2). For *d2* = 1, there are {8, 4, 4} = 16 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 6, 7}. For *d2* = 2, there are {7, 3, 4} = 14 ways to assign *dfi*, *dfn*, and *dfd* for *n2* = {5, 6, 7}. There are 16 + 14 = 30 ways to assign *i1*, *d1, d2, d3, n2*, and *n5* when *n1* = 8 and 2! = 2 ways to assign *n3* and *n4*, giving 30 * 2 = 60 possibilities.

The search space for single-digit *i* = 8 + 44 + 128 + 104 + 88 + 26 + 60 = 458. This is a reduction of 99.9% for this case.

**Two-digit i, cases for i1:**

**2:** For 21 <= *i* <= 28, then 72 <= *n/d* <= 79 and the maximum value for *d* is the greater of 9764/72 = 135 (where *i2* = 8) and 9864/73 = 135 (where *i2* = 7). This makes *d1* = 1 and *d2* = 3, giving a range for *d* of 134 to 134 (since *d3* can’t be 5). Since 9876/134 = 73, the minimum value for *i* is 100 – 73 = 27, making *i2* = 7, 8 or 9. There is no way to assign *dfi*, *dfn*, and *dfd* where *dfd* = 4 and *dfi* = 7, 8 or 9, and *dfn* is not 1, 2 or 3. So *i1* can’t be 2.

**3:** For 31 <= *i* <= 38, then 62 <= *n/d* <= 69 and the maximum value for *d* is the greater of 9764/62 = 157 (where *i2* = 8) and 9864/63 = 156 (where *i2* = 7). This gives a range for *d* of 124 to 157, and makes *d1* = 1 and *d2* = 2, 4 or 5. If *d1* = 1, then *i2* can’t be 1. If *i2* is 2, then *n/d* = 68 and the minimum value for *d* is 146 (since *i* is 32 and *d3* can’t be 5). Since 146 * 68 > 9876, *i2* can’t be 2, making *i2* = 4, 6, 7 or 8. The minimum value for *n* is 124 * 62 = 7688, making *n1* = 7, 8 or 9. There are 11 combinations for *dfi*, *dfn*, and *dfd* where *dfi* = 4, 6, 7 or 8 and that don’t have a 1 or 3: (4,2,7), (6,2,8), (6,8,2), (6,8,7), (7,2,4), (7,4,8), (7,6,2), (7,8,6), (8,2,6), (8,4,2) and (8,4,7). For *d2* = 2, the minimum value for *n* is 124 * 62 = 7688 and there are {0, 0, 4} = 4 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {7, 8, 9}. For *d2* = 4, the minimum value for *n* is 142 * 62 = 8804 and there are {1, 6} = 7 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {8, 9}. For *d2* = 5, the minimum value for *n* is 152 * 62 = 9424 and there are 11 ways to assign *dfi*, *dfn*, and *dfd* when *n1* = 9. There are 4 + 7 + 11 = 22 ways to assign *i2*, *d1, d2, d3, n1*, and *n4* when *i1* = 3 and 2! = 2 ways to assign *n2* and *n3*, giving 22 * 2 = 44 possibilities.

**4:** For 41 <= *i* <= 48, then 52 <= *n/d* <= 59 and the maximum value for *d* is the greater of 9763/52 = 187 (where *i2* = 8) and 9863/53 = 186 (where *i2* = 7). This gives a range for *d* of 123 to 187, and makes *d1* = 1 and *d2* = 2, 3, 5, 6, 7 or 8. There are 14 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 1 or 4: (2,6,7), (2,8,6), (3,2,6), (3,6,8), (3,9,7), (6,2,3), (6,2,8), (6,8,2), (6,8,7), (7,6,2), (7,8,6), (7,9,3), (8,2,6) and (8,6,3). For *d2* = 2, the minimum value for *n* is 123 * 52 = 6396 and there are {2, 2, 2, 4} = 10 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For *d2* = 3, the minimum value for *n* is 132 * 52 = 6864 and there are {0, 4, 2, 8} = 14 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For *d2* = 5, the minimum value for *n* is 152 * 52 = 7904 and there are {8, 6, 12} = 26 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {7, 8, 9}. For *d2* = 6, the minimum value for *n* is 162 * 52 = 8424 and there are {2, 0} = 2 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {8, 9}. For *d2* = 7, the minimum value for *n* is 172 * 52 = 8424, but *n1* can’t be 8 since *i2* is 8 and 172 * 53 = 9116, so *n1* = 9 and there are 6 ways to assign *dfi*, *dfn*, and *dfd*. For *d2* = 8, the minimum value for *n* is 182 * 53 = 9696, but since *i2* = 7, the largest value for *n* is 9653; if *i2* = 6 then 182 * 54 = 9828, but the largest value for *n* is 9753, so *d2* can’t be 8. There are 10 + 14 + 26 + 2 + 6 = 58 ways to assign *i2*, *d1, d2, d3, n1*, and *n4* when *i1* = 4 and 2! = 2 ways to assign *n2* and *n3*, giving 58 * 2 = 116 possibilities.

**5:** For 51 <= *i* <= 58, then 42 <= *n/d* <= 49 and the maximum value for *d* is the greater of 9764/42 = 232 (where *i2* = 8) and 9864/43 = 229 (where *i2* = 7). This gives a range for *d* of 123 to 218 (because *dfd* can’t be 0, 1 or 9), and makes (*d1, d2)* = (1, 2), (1, 3), (1, 4), (1, 6), (1, 7), (1, 8), (1, 9) or (2, 1). None of the combinations for *dfi*, *dfn*, and *dfd* have a 5, so all 36 are possible. For (*d1, d2)* = (1, 2) the minimum value for *n* is 123 * 42 = 5166, but *n1* can’t be 5, and there are {6, 4, 2, 8} = 20 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For (*d1, d2)* = (1, 3) the minimum value for *n* is 132 * 42 = 5544, but *n1* can’t be 5, and there are {7, 6, 4, 14} = 31 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For (*d1, d2)* = (1, 4) the minimum value for *n* is 142 * 42 = 5964, but *n1* can’t be 5, and there are {2, 8, 6, 12} = 28 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For (*d1, d2)* = (1, 6) the minimum value for *n* is 162 * 42 = 6804, but *n1* can’t be 6, and there are {6, 6, 10} = 32 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {7, 8, 9}. For (*d1, d2)* = (1, 7) the minimum value for *n* is 172 * 42 = 7224, but *n1* can’t be 7, and there are {4, 14} = 18 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {8, 9}. For (*d1, d2)* = (1, 8) the minimum value for *n* is 182 * 43 = 7826, but *n1* can’t be 7 because *i2* = 7 and 182 * 44 = 8008, so *n1* = 9 and there are 8 ways to assign *dfi*, *dfn*, and *dfd*. For (*d1, d2)* = (1, 9) the minimum value for *n* is 192 * 43 = 8256, so *n1* = 8 and there are 8 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 7, 8, 9}. For (*d1, d2)* = (2, 1) the minimum value for *n* is 213 * 42 = 8946, but *n1* can’t be 8 because *i2* = 8, so *n1* = 9 and there are 8 ways to assign *dfi*, *dfn*, and *dfd*. There are 20 + 31 + 28 + 32 + 18 + 8 + 8 + 8 = 153 ways to assign *i2*, *d1, d2, d3, n1*, and *n4* when *i1* = 5 and 2! = 2 ways to assign *n2* and *n3*, giving 153 * 2 = 306 possibilities.

**6:** For 61 <= *i* <= 68, then 32 <= *n/d* <= 39 and the maximum value for *d* is the greater of 9754/32 = 304 (where *i2* = 8) and 9864/33 = 299 (where *i2* = 7). This gives a range for *d* of 123 to 298 (*d2* can’t be zero), and makes *d1* = 1 or 2. There are 16 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 6 or don’t have both a 1 and 2: (1,3,7), (1,7,3), (2,4,3), (2,4,8), (3,4,2), (3,7,1), (3,8,4), (3,9,7), (4,2,7), (4,8,3), (7,2,4), (7,3,1), (7,4,8), (7,9,3), (8,4,2) and (8,4,7). For *d1* = 1 the minimum value for *n* is 126 * 32 = 4032 (because 123 * 32 = 3936, but *n1* can’t be 3 if *d3* = 3; 124 * 32 = 3968, but *n3* can’t be 6; 124 * 33 = 4092; and *d3* can’t be 5), and there are {2, 12, 6, 6, 10} = 36 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {4, 5, 7, 8, 9}. For *d1* = 2 the minimum value for *n* is 213 * 32 = 6816, but *n1* can’t be 6, and there are {2, 6, 8} = 16 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {7, 8, 9}. There are 36 + 16 = 52 ways to assign *i2*, *d1, d3, n1*, and *n4* when *i1* = 6 and 3! = 6 ways to assign *d2*, *n2* and *n3*, giving 52 * 6 = 312 possibilities.

**7:** For 71 <= *i* <= 78, then 22 <= *n/d* <= 29 and the maximum value for *d* is 9654/22 = 438. This gives a range for *d* of 123 to 436 (*dfd* can’t be 7 or 8 when *i* = 78), and makes *d1* = 1, 2, 3 or 4. There are 22 combinations for *dfi*, *dfn*, and *dfd* that don’t have a 7: (1,2,8), (1,4,6), (1,6,4), (1,8,2), (2,4,3), (2,4,8), (2,8,1), (2,8,6), (3,2,6), (3,4,2), (3,6,8), (3,8,4), (4,6,1), (4,8,3), (6,2,3), (6,2,8), (6,4,1), (6,8,2), (8,2,1), (8,2,6), (8,4,2) and (8,6,3). For *d1* = 1 the minimum value for *n1* = 3 (because 134 * 22 = 2948, but *n3* can’t be 4 if *d3* = 4; 136 * 22 = 2992, but *n2* and *n3* can’t both be 9), and there are {6, 8, 14, 6, 4, 2, 14} = 54 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {3, 4, 5, 6, 8, 9}. For *d1* = 2 the minimum value for *n* is 213 * 22 = 4686, and there are {1, 8, 2, 4, 8} = 23 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {4, 5, 6, 8, 9}. For *d1* = 3 the minimum value for *n* is 312 * 22 = 6864, and there are {6, 4, 14} = 24 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {6, 8, 9}. For *d1* = 4 the minimum value for *n* is 412 * 22 = 9064, and there are 12 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = 9. There are 54 + 23 + 24 + 12 = 113 ways to assign *i2*, *d1, d3, n1*, and *n4* when *i1* = 7 and 3! = 6 ways to assign *d2*, *n2* and *n3*, giving 113 * 6 = 678 possibilities.

**8:** For 81 <= *i* <= 87, then 13 <= *n/d* <= 19 and the maximum value for *d* is 9654/13 = 742. This gives a range for *d* of 123 to 742, and makes *d1* = 1, 2, 3, 4, 5, 6 or 7. There are 18 combinations for *dfi*, *dfn*, and *dfd* that don’t have an 8: (1,3,7), (1,4,6), (1,6,4), (1,7,3), (2,4,3), (2,6,7), (3,2,6), (3,4,2), (3,7,1), (3,9,7), (4,2,7), (4,6,1), (6,2,3), (6,4,1), (7,2,4), (7,3,1), (7,6,2) and (7,9,3). For *d1* = 1 the minimum value for *n1* is 2 (because 1976/13 = 152, so if *d1* = 1, then *n1* <= 1), and there are {2, 4, 6, 10, 6, 4, 8} = 40 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {2, 3, 4, 5, 6, 7, 9}. For *d1* = 2 the minimum value for *n1* is 3 (because 213 * 13 = 2769, but *n1* can’t be 2 if *d1* = 2), and there are {4, 6, 10, 6, 4, 8} = 38 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {3, 4, 5, 6, 7, 9}. For *d1* = 3 the minimum value for *n* is 312 * 13 = 4056, and there are {2, 8, 3, 4, 8} = 35 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {4, 5, 6, 7, 9}. For *d1* = 4 the minimum value for *n* is 412 * 13 = 5356, and there are {10, 6, 2, 9} = 27 ways to assign *dfi*, *dfn*, and *dfd* for *n1* = {5, 6, 7, 9}. There are 40 + 38 + 35 + 27 = 140 ways to assign *i2*, *d1, d3, n1*, and *n4* when *i1* = 8 and 3! = 6 ways to assign *d2*, *n2* and *n3*, giving 140 * 6 = 840 possibilities.

**9:** For 91 <= *i* <= 98, then 2 <= *n/d* <= 9. For this range it makes more sense to look at each value for *i2* = {1, 2, 3, 4, 6, 7, 8} separately.

*i2* = 1, *n/d* = 9: In this case *d1* > *n1* and the combinations of *dfi*, *dfn* and *dfd* where *dfi* = 1 that don’t have a 9 are (1,2,8), (1,3,7), (1,4,6), (1,6,4), (1,7,3) and (1,8,2). If *n1* = 2, the maximum value for *d* = 2876/9 = 319, but *d2* can’t be 0 or 1, so *n1* can’t be 2. If *n1* = 3, the maximum value for *d* = 3876/9 = 430, which gives a range for *d* of 426 to 428 (because *dfd* can’t be 0, 1, 4, 5 or 9), but there are no combinations for *dfn* and *dfd* that don’t have a 2, 3 or 4, so *n1* can’t be 3. If *n1* = 4, the maximum value for *d* = 4876/9 = 541, which gives a range for *d* of 523 <= *d* <= 538 (because *dfd* can’t be 0, 1 or 9), and there are {2, 2} = 4 ways to assign *dfn* and *dfd* for *d2* = {2, 3}. If *n1* = 5, the maximum value for *d* = 5876/9 = 652, which gives a range for *d* of 623 <= *d* <= 648 (because *d2* can’t be 5 and *dfd* can’t be 9), and there are {2, 2, 4} = 8 ways to assign *dfn* and *dfd* for *d2* = {2, 3, 4}. If *n1* = 6, the maximum value for *d* = 6874/9 = 763, which gives a range for *d* of 723 <= *d* <= 758 (because *d2* can’t be 6 and *dfd* can’t be 9), and there are {0, 2, 2, 2} = 6 ways to assign *dfn* and *dfd* for *d2* = {2, 3, 4, 5}. If *n1* = 7, the maximum value for *d* = 7864/9 = 873, which gives a range for *d* of 823 <= *d* <= 867 (because *d2* can’t be 7 and *dfd* can’t be 9), and there are {2, 2, 0, 2, 0} = 6 ways to assign *dfn* and *dfd* for *d2* = {2, 3, 4, 5, 6}. *n1* can’t be 8 because *d1* can’t be 9. There are 4 + 8 + 6 + 6 = 24 ways to assign *d1, d2, d3, n1*, and *n4* when *i* = 91 and 2! = 2 ways to assign *n2* and *n3*, giving 24 * 2 = 48 possibilities.

*i2* = 2, *n/d* = 8: The combinations of *dfi*, *dfn* and *dfd* where *dfi* = 2 that don’t have a 9 are (2,4,3), (2,4,8), (2,6,7), (2,8,1) and (2,8,6). If *n1* = 1, the maximum value for *d* = 1876/8 = 234, but *d1* can’t be 2, so *n1* can’t be 1. If *n1* = 3, the maximum value for *d* = 3876/8 = 484, but *n2* and *d2* can’t both be 8 and 480 * 8 = 3840, *d2* can’t be 8 when *n1* = 3, which gives a range for *d* of 416 <= *d* <= 478, and there are {2, 3, 1, 2} = 8 ways to assign *dfn* and *dfd* for *d2* = {1, 5, 6, 7}. If *n1* = 4, the minimum value for *d* = 4136/8 = 518 and the maximum value for *d* = 4876/8 = 609, which gives a range for *d* of 518 <= *d* <= 587 (because *d2* can’t be 0 or 9 and *dfd* can’t be 9), and there are {0, 3, 1, 2, 1} = 7 ways to assign *dfn* and *dfd* for *d2* = {1, 3, 6, 7, 8}. If *n1* = 5, the minimum value for *d* = 5134/8 = 642 and the maximum value for *d* = 5876/8 = 734, but *n3* and *d1* can’t both be 7, so the maximum value for *d* = 5764/8 = 733, which gives a range for *d* of 643 <= *d* <= 731, and there are {0, 3, 1, 3, 1} = 8 ways to assign *dfn* and *dfd* for (*d1*, *d2*) = {(6,4), (6,7), (6,8), (7,1), (7,3)}. If *n1* = 6, the minimum value for *d* = 6134/8 = 767 and the maximum value for *d* = 6874/8 = 859, but *n2* and *d1* can’t both be 8, so the maximum value for *d* = 6754/8 = 844, which gives a range for *d* of 781 <= *d* <= 843, and there are {1, 1, 0, 0} = 2 ways to assign *dfn* and *dfd* for (*d1*, *d2*) = {(7,8), (8,1), (8,3), (8,4)}. If *n1* = 7, the minimum value for *d* = 7134/8 = 892, which is larger than 864, the largest possible value for *d*, so *n1* < 7. There are 8 + 7 + 8 + 2 = 25 ways to assign *d1, d2, d3, n1*, and *n4* when *i* = 92 and 2! = 2 ways to assign *n2* and *n3*, giving 25 * 2 = 50 possibilities.

*i2* = 3, *n/d* = 7: The combinations of *dfi*, *dfn* and *dfd* where *dfi* = 3 that don’t have a 9 are (3,2,6), (3,4,2), (3,6,8), (3,7,1) and (3,8,4). If *n1* = 1, the maximum value for *d* = 1876/7 = 268, which gives a range for *d* of 246 to 268, and there are {1, 2, 1} = 4 ways to assign *dfn* and *dfd* for *d2* = {4, 5, 6}. If *n1* = 2, the minimum value for *d* = 2146/7 = 307 and the maximum value for *d* = 2876/7 = 410, and *d1* can’t be 3, so *n1* can’t be 2. If *n1* = 4, the minimum value for *d* = 4126/7 = 590 and the maximum value for *d* = 4876/7 = 696, which gives a range for *d* of 612 <= *d* <= 687 (because *d2* can’t be 0 or 9 and *dfd* can’t be 9), and there are {0, 1, 1, 0, 1} = 3 ways to assign *dfn* and *dfd* for *d2* = {1, 2, 5, 6, 7, 8}. If *n1* = 5, the minimum value for *d* = 5124/7 = 732 and the maximum value for *d* = 5876/7 = 839, which gives a range for *d* of 741 <= *d* <= 827 (because *d2* can’t be 3 and *dfd* can’t be 9), and there are {2, 2, 2, 2, 1} = 9 ways to assign *dfn* and *dfd* for (*d1*, *d2*) = {(7,4), (7,6), (7,8), (8,1), (8,2)}. If *n1* = 6, the minimum value for *d* = 6124/7 = 875, which is greater than the largest available value for *d* of 874, so *n1* < 6. There are 4 + 3 + 9 = 16 ways to assign *d1, d2, d3, n1*, and *n4* when *i* = 93 and 2! = 2 ways to assign *n2* and *n3*, giving 16 * 2 = 32 possibilities.

* i2* = 4,

*n/d*= 6: The combinations of

*dfi*,

*dfn*and

*dfd*where

*dfi*= 4 that don’t have a 9 are (4,2,7), (4,6,1) and (4,8,3). If

*n1*= 1,

*dfd*= 3 or 7 and the maximum value for

*d*= 1876/6 = 312, which gives a range for

*d*of 237 to 287, and there are {0, 1, 1, 1, 0} = 3 ways to assign

*dfn*and

*dfd*for

*d2*= {3, 5, 6, 7, 8}. If

*n1*= 2, the minimum value for

*d*= 2136/6 = 356 and the maximum value for

*d*= 2876/6 = 479, and

*d1*can’t be 3 or 4, so

*n1*can’t be 2. If

*n1*= 3, the minimum value for

*d*= 3126/6 = 521 and the maximum value for

*d*= 3876/6 = 646, which gives a range for

*d*of 521 <=

*d*<= 628 (because

*d2*can’t be 3 or 4 and

*dfd*can’t be 9), and there are {1, 1, 1, 2, 1, 0} = 6 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(5,2), (5,6), (5,7), (5,8), (6,1), (6,2)}. If

*n1*= 5, the minimum value for

*d*= 5123/6 = 854 and the maximum value for

*d*= 5876/6 = 979, which gives a range for

*d*of 861 <=

*d*<= 876 (because

*d2*can’t be 5 or 9, and

*dfd*can’t be 9), and there are {1, 1} = 2 ways to assign

*dfn*and

*dfd*for

*d2*= {6, 7}.

*n1*<

*n/d*, so

*n1*< 6. There are 3 + 6 + 2 = 11 ways to assign

*d1, d2, d3, n1*, and

*n4*when

*i*= 94 and 2! = 2 ways to assign

*n2*and

*n3*, giving 11 * 2 = 22 possibilities.

*= 6,*

i2

i2

*n/d*= 4: The combinations of

*dfi*,

*dfn*and

*dfd*where

*dfi*= 6 that don’t have a 9 are (6,2,3), (6,2,8), (6,4,1), (6,8,2) and (6,8,7). If

*n1*= 1, the minimum value for d = 1234/4 = 309 and the maximum value for

*d*= 1876/4 = 469, which gives a range for

*d*of 324 to 458 (because

*d2*can’t be 1 or 6 and

*dfd*can’t be 9), and there are {1, 3, 3, 2, 0, 1, 3, 4} = 17 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(3,2), (3,4), (3,5), (3,7), (3,8), (4,2), (4,3), (4,5)}. If

*n1*= 2, the minimum value for

*d*= 2134/4 = 534 and the maximum value for

*d*= 2874/4 = 718, which gives a range for

*d*of 534 to 718, and there are {2, 1, 1, 1, 0} = 5 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(5,3), (5,4), (5,7), (5,8), (7,1)}. If

*n1*= 3, the minimum value for

*d*= 3124/4 = 781 and the maximum value for

*d*= 3874/4 = 968, which gives a range for

*d*of 781 <=

*d*<= 874 (because no digits in

*d*can be 9), and there are {1, 0, 1, 0, 1, 1} = 4 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(7,8), (8,1), (8,2), (8,4), (8,5), (8,7)}.

*n1*<

*n/d*, so

*n1*< 4. There are 17 + 5 + 4 = 26 ways to assign

*d1, d2, d3, n1*, and

*n4*when

*i*= 96 and 2! = 2 ways to assign

*n2*and

*n3*, giving 26 * 2 = 52 possibilities.

*= 7,*

i2

i2

*n/d*= 3: The combinations of

*dfi*,

*dfn*and

*dfd*where

*dfi*= 7 that don’t have a 9 are (7,2,4), (7,3,1), (7,4,8), (7,6,2) and (7,8,6). If

*n1*= 1, the minimum value for d = 1234/3 = 412 and the maximum value for

*d*= 1864/3 = 621, which gives a range for

*d*of 423 to 586 (because no digits in

*d*can be 0, 1 or 9, and

*dfd*can’t be 9), and there are {1, 2, 0, 1, 2, 4, 2, 2, 2} = 16 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(4,2), (4,5), (4,6), (4,8), (5,2), (5,3), (5,4), (5,6), (5,8)}. If

*n1*= 2, the minimum value for

*d*= 2134/3 = 712 and the maximum value for

*d*= 2864/3 = 954, which gives a range for

*d*of 813 to 864 (because no digits in

*d*can be 7 or 9), and there are {0, 0, 1, 1, 1} = 3 ways to assign

*dfn*and

*dfd*for

*d2*= {1, 3, 4, 5, 6}.

*n1*<

*n/d*, so

*n1*< 3. There are 16 + 3 = 19 ways to assign

*d1, d2, d3, n1*, and

*n4*when

*i*= 97 and 2! = 2 ways to assign

*n2*and

*n3*, giving 19 * 2 = 38 possibilities.

*= 8,*

i2

i2

*n/d*= 2: Since

*n/d*= 2,

*n1*= 1. The combinations of

*dfi*,

*dfn*and

*dfd*where

*dfi*= 8 that don’t have a 1 or 9 are (8,2,6), (8,4,2), (8,4,7) and (8,6,3). The minimum value for d = 1234/2 = 617 and the maximum value for

*d*= 1764/2 = 882, which gives a range for

*d*of 623 to 764 (because no digits in

*d*can be 0, 1, 8 or 9, and

*dfd*can’t be 9), and there are {1, 2, 0, 2, 1, 1, 2, 2, 3, 1} = 15 ways to assign

*dfn*and

*dfd*for (

*d1*,

*d2*) = {(6,2), (6,3), (6,4), (6,5), (6,7), (7,2), (7,3), (7,4), (7,5), (7,6)}. There are 15 ways to assign

*d1, d2, d3, n1*, and

*n4*when

*i*= 98 and 2! = 2 ways to assign

*n2*and

*n3*, giving 15 * 2 = 30 possibilities.

For 91 <= *i* <= 98, there are 48 + 50 + 32 + 22 + 52 + 38 + 30 = 272 possibilities.

The search space for two-digit *i* = 44 + 116 + 306 + 312 + 768 + 840 + 272 = 2658. This is a reduction of 99.3% for this case.

**Final tally**

The final total search space is 458 + 2658 = 3116 possible permutations of the digits 1 to 9 that produce a mixed fraction that equals 100.

By pruning, the search space is reduced by over 99.57%. Of the 7 single-digit and 14 two-digit sub-spaces, 5 contained a result (single-digit *i* = 3, two-digit 81 <= *i* <= 87, *i* = 91, *i* = 94, and *i* = 96). Those sub-spaces contained 44 + 840 + 48 + 22 + 52 = 996 possibilities containing 1 + 3 + 3 + 1 + 3 = 11 results.

### Your Answer

## More puzzles to try-

### What is the logic behind these ?

3 + 3 = 3 5 + 4 = 4 1 + 0 = 3 2 + 3 = 4 ...Read More »### Five greedy pirates and gold coin distribution Puzzle

Five puzzleFry ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all ...Read More »### Defective stack of coins puzzle

There are 10 stacks of 10 coins each. Each coin weights 10 gms. However, one stack of coins is defective ...Read More »### Dr. Willam and surgical gloves puzzle

Dr.Willam wants to operate for three different persons who were wounded. But he had only two surgical gloves. There is ...Read More »### Which clock works best?

Which clock works best? The one that loses a minute a day or the one that doesn’t work at all?Read More »### To which one will the needle touch 1 OR 2

In the gear arrangement, To which one will the needle touch, 1 OR 2Read More »### How did he know?

A man leaves his house in the morning to go to office and kisses his wife. In the evening on ...Read More »### Which letter replaces the question mark

Which letter replaces the question markRead More »### Overthrow your math teacher!

Some numbers are very mysterious. They are all integers. They have more than one digit. If you multiply them with ...Read More »### Crossing the Bridge Puzzle (Bridge and torch problem)

A bridge will collapse in 17 minutes. 4 people want to cross it before it will collapse. It is a ...Read More »### Probability of having boy

In a country where everyone wants a boy, each family continues having babies till they have a boy. After some ...Read More »### Magical flowers!!

A devotee goes to three temples, temple1, temple2 and temple3 one after the other. In front of each temple, there ...Read More »### How many times a day do all three hands of analog watch overlaps?

There is a perfectly accurate analog watch with hour, minute, and second hands. How many times a day do all three ...Read More »### Tuesday, Thursday what are other two days staring with T?

Four days are there which start with the letter ‘T‘. I can remember only two of them as “Tuesday , Thursday”. ...Read More »### How could only 3 apples left

Two fathers took their sons to a fruit stall. Each man and son bought an apple, But when they returned ...Read More »### How Many Eggs ?

A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over ...Read More »### Minimum number of persons needed to cross a Desert

In the middle of the confounded desert, there is the lost city of “Ash”. To reach it, you will have ...Read More »### Lateral thinking sequence Puzzle

Solve this logic sequence puzzle by the correct digit- 8080 = 6 1357 = 0 2022 = 1 1999 = ...Read More »### Pizza Cost Math Brain Teaser

Jasmine, Thibault, and Noah were having a night out and decided to order a pizza for $10. It turned out ...Read More »### Which room is safest puzzle

A murderer is condemned to death. He has to choose between three rooms. The first is full of raging fires, ...Read More »