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27 Aug 2020, 08:00
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Difficulty:
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When the digits of two-digit positive integer m are reversed, the result is two-digit positive integer n. What is the largest integer that must be a factor of 4m - 7n?
A. 1
B. 9
C. 11
D. 33
E. 99
A. 1
B. 9
C. 11
D. 33
E. 99
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Updated on: 23 Sep 2020, 12:32
Originally posted by Varunsawhney8 on 27 Aug 2020, 08:07.
Last edited by Varunsawhney8 on 23 Sep 2020, 12:32, edited 1 time in total.
Last edited by Varunsawhney8 on 23 Sep 2020, 12:32, edited 1 time in total.
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When the digits of two-digit positive integer m are reversed, the result is two-digit positive integer n. What is the largest integer that must be a factor of 4m - 7n?
Let 2 digit positve number m= ab
Reverse of m = n = ba
We know m=10a+b and n=10b+a
in the exp.
4m-7a
=40a+4b-(70b+7a)
=33a-66b
taking 33 common
=33(a-2b)
So largest integer is 33
Answer: D
Posted from my mobile device
Let 2 digit positve number m= ab
Reverse of m = n = ba
We know m=10a+b and n=10b+a
in the exp.
4m-7a
=40a+4b-(70b+7a)
=33a-66b
taking 33 common
=33(a-2b)
So largest integer is 33
Answer: D
Posted from my mobile device
General Discussion
27 Aug 2020, 08:09
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One way to represent a 2-digit number is 10x+y in which x and y are the 2 digits.
For ex: 45 can be expanded as 10*4+5 and its reverse 54 can be represented as 10*5 + 4
So, let's assume m is equal to 10x+y then n would be equal to 10y+x
4m - 7n would become 4(10x+y) - 7(10y+x)
=> 40x + 4y - 70y - 7x
=> 33x - 66y
=> 33(x - 2y)
Therefore, the largest factor would be 33
Hence D is the answer
For ex: 45 can be expanded as 10*4+5 and its reverse 54 can be represented as 10*5 + 4
So, let's assume m is equal to 10x+y then n would be equal to 10y+x
4m - 7n would become 4(10x+y) - 7(10y+x)
=> 40x + 4y - 70y - 7x
=> 33x - 66y
=> 33(x - 2y)
Therefore, the largest factor would be 33
Hence D is the answer
