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07 Dec 2021, 05:30
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (01:52) correct
50%
(01:56)
wrong
based on 92
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A product formed of a two digit number and the two digit number formed by reversing the digits of the first two digit number results in a number that ends in the digit 5. How many such two digit numbers are possible?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Are You Up For the Challenge: 700 Level Questions
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Are You Up For the Challenge: 700 Level Questions
07 Dec 2021, 05:38
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If we take a two digit number AB and reverse its digits to get BA, and multiply AB*BA, the units digit we get will just be the units digit of B*A. So we need a two digit number AB where A*B will end in 5, so one of the two digits needs to be 5, and the other needs to be odd (if the other digit is even, the product of the digits will end in 0). So our number can be 51, 53, 55, 57 or 59 if the first digit is 5, or it can be 15, 35, 75 or 95 if the second digit is 5 (here we want to be sure not to count '55' twice), and the answer is nine.
Updated on: 07 Dec 2021, 23:21
Originally posted by sanjitscorps18 on 07 Dec 2021, 05:47.
Last edited by sanjitscorps18 on 07 Dec 2021, 23:21, edited 1 time in total.
Last edited by sanjitscorps18 on 07 Dec 2021, 23:21, edited 1 time in total.
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As per the question we have
(10x + y)(10y +x) = abc5 or ab5
=> 100xy + 10(x^2 + y^2) + xy = abc5 or ab5
=> 101xy + 10(x^2 + y^2) = abc5 or ab5
Now 10(x^2 + y^2) cannot end in 5 as a units digit hence we know that 101 * x * y would have units digit as 5. This is possible if x * y gives 5 in the units place as we have 1 in the units place of 101.
The possible combinations would be created when any odd integer is multiplied by 5. Listing the (x, y) combinations below
(1, 5), (3, 5), (5, 5), (7, 5), (9, 5), (5, 1), (5, 3), (5, 7), (5, 9)
Hence 9 combinations
Option E
(10x + y)(10y +x) = abc5 or ab5
=> 100xy + 10(x^2 + y^2) + xy = abc5 or ab5
=> 101xy + 10(x^2 + y^2) = abc5 or ab5
Now 10(x^2 + y^2) cannot end in 5 as a units digit hence we know that 101 * x * y would have units digit as 5. This is possible if x * y gives 5 in the units place as we have 1 in the units place of 101.
The possible combinations would be created when any odd integer is multiplied by 5. Listing the (x, y) combinations below
(1, 5), (3, 5), (5, 5), (7, 5), (9, 5), (5, 1), (5, 3), (5, 7), (5, 9)
Hence 9 combinations
Option E
07 Dec 2021, 06:38
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sanjitscorps18
I think everything in your solution was correct to this point, but I think you changed perspectives on what your "xy" means in the last line of your solution. Throughout your solution, you were treating x and y as single-digit numbers, and when you multiplied everything out, "xy" became the product of two single-digit numbers. But in the last line, where you list the valid values of xy, I think your "xy" became something else -- it's not the product of x and y any more, and instead it became a two-digit number where x is the tens digit and y is the units digit. But that's not what your "xy" represents, and at the end, you're looking for two one-digit numbers x and y which have a product ending in 5, so 25 and 45, for example, are not possible values of xy, and 51 and 53 are.
