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A palindrome, such as 12321, is a number that remains the sa

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A palindrome, such as 12321, is a number that remains the sa

by Max@Math Revolution » Mon Jul 01, 2019 12:23 am

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[GMAT math practice question]

A palindrome, such as 12321, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?

A. 18
B. 24
C. 28
D. 32
E. 36
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by GMATGuruNY » Mon Jul 01, 2019 1:49 am
Max@Math Revolution wrote:[GMAT math practice question]

A palindrome, such as 12321, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?

A. 18
B. 24
C. 28
D. 32
E. 36
The difference between the two palindromes = 32.
Since subtracting 32 from the 4-digit palindrome must yield a 3-digit palindrome. the 4-digit palindrome must be just a bit more than 1000.
Test 1001:
1001 - 32 = 969
Success!
Subtracting 32 from 1001 (a 4-digit palindrome) yields 969 (a 3-digit palindrome).
Since x=969, the sum of its digits = 9+6+9 = 24.

The correct answer is B.
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by Max@Math Revolution » Tue Jul 02, 2019 11:42 pm
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Let x = ABA and x + 32 = CDDC.
Then CDDC = ABA + 32
Since CDDC has a thousands digit, we must have A = 9.
Then the units digit of CDDC is equal to the units digit of A + 2 = 11.
And, looking at the tens digits, we must have B + 3 + 1 ≥ 10.
Therefore, B ≥ 6.
The possible values for B are:
B = 6, 7, 8, 9.
Let's check which value gives a palindrome for ABA + 32:
969 + 32 = 1001
979 + 32 = 1011
989 + 32 = 1021
999 + 32 = 1031
The only palindrome is 1001, so x = 969.
Thus, the sum of the digits of x is 9 + 6 + 9 = 24.

Therefore, B is the answer.
Answer: B
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