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A product formed of a two digit number and the two digit number formed 07 Dec 2021, 05:30
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49% (01:54) correct 51% (01:58) wrong based on 89 sessions

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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


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E
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A product formed of a two digit number and the two digit number formed 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.
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A product formed of a two digit number and the two digit number formed 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.
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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
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A product formed of a two digit number and the two digit number formed 07 Dec 2021, 06:38
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sanjitscorps18

Since xy is multiplied by 1 in the units place Now combinations of xy that can result in 5 as a units digits would be

15, 25, 35,......95
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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.
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A product formed of a two digit number and the two digit number formed 07 Dec 2021, 08:05
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sanjitscorps18

Since xy is multiplied by 1 in the units place Now combinations of xy that can result in 5 as a units digits would be

15, 25, 35,......95

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.
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I get your point and I could have been clearer. I based this solution on the premise that the actual number is 10x + y. This order of the number is important to avoid dual counting. Now basis this scenario in all my equations I'm treating x and y together as a product. Since I keep the original number as 10x + y hence I'm only considering y values as 5 and therefore we only have 9 combinations. If the order is skipped then of course I agree to your suggestion that it could have been more combinations with 5 as units digits. Hope this makes sense.
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A product formed of a two digit number and the two digit number formed 07 Dec 2021, 09:37
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sanjitscorps18

If the order is skipped then of course I agree to your suggestion that it could have been more combinations with 5 as units digits. Hope this makes sense.
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It's just because your list of possible values included 25, and the number cannot be 25 here (because 25*52 does not end in 5), that I looked over your solution to work out how you might have arrived at that value. If, in the last line of your solution, you were thinking of 'xy' as if it were a two-digit number ending in 'y', but elsewhere you were treating 'xy' as a product of two single-digit numbers, that would explain why you concluded 25 was a possible value, but perhaps I've misunderstood what you were doing.
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A product formed of a two digit number and the two digit number formed 07 Dec 2021, 23:27
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sanjitscorps18

If the order is skipped then of course I agree to your suggestion that it could have been more combinations with 5 as units digits. Hope this makes sense.

It's just because your list of possible values included 25, and the number cannot be 25 here (because 25*52 does not end in 5), that I looked over your solution to work out how you might have arrived at that value. If, in the last line of your solution, you were thinking of 'xy' as if it were a two-digit number ending in 'y', but elsewhere you were treating 'xy' as a product of two single-digit numbers, that would explain why you concluded 25 was a possible value, but perhaps I've misunderstood what you were doing.
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Hey Ian, I've edited my answer. Hope it's clear now and comprehensible for the forum members. Thanks.
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