Five digit number of which First and last digits are same riddle
I am thinking of a five digit number such that:
First and last digits are same, their submission is an even number and multiplication is an odd number and is equal to the fourth number. Subtract five from it and we obtain the second number. Then divide into exact half and we get the 3rd number.
Which number I am thinking of?
The number is 34293
Deduction:
Assumptions: The terms ‘fourth number‘, ‘second number‘, ‘third number‘ etc, refer to the respective digits of the number.
Conditions given:
1. First and last digits are same.
2. Their submission is an even number
3. [Their] multiplication is an odd number
4. [That odd number] is equal to the fourth number.
5. Subtract five from it [the fourth digit] and we obtain the second number.
6. Then divide into exact half and we get the 3rd number.
From Statements 1 and 3, it follows that the first and last digits (which are same) MUST be odd, because their product is odd.
So, the possible digits are 1, 3, 5, 7 and 9. — (1)
Now, Statement 3 says that the product mentioned in S3 is the fourth digit, which means that the product must be a single digit.
Hence, This cuts down the possible digits listed in (1) to 1 and 3 only. (2)
Now, S5 asks to subtract five (from the 4th digit). With this, the possibility of ‘1’ is ruled out.
Hence, it follows that the first and fifth digits must be 3. (3)
From (1), (2) and (3), the number is in the form 3XYZ3.
Now, as per S4, Z = product of first & last digits. Hence, the number is 3XY93. (4)
S5 gives the second digit as 9 – 5 = 4 => Number = 34Y93 (5)
Finally, the 6th statement gives the remaining digit as half of the second number, ie., half of 4 => 2.
Hence, the result.
Your Answer
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