If r is the remainder when ab is divided by ba, where both ab and ba

Question

If  r  is the remainder when  ab  is divided by  ba , where  ab  and  ba  are positive two-digit integers with  a > b , what is the maximum possible value of  r ?

A. 9
B. 18
C. 27
D. 36
E. 45

Bunuel · Accepted Answer

If  r  is the remainder when  ab  is divided by  ba , where  ab  and  ba  are positive two-digit integers with  a > b , what is the maximum possible value of  r ?

A. 9
B. 18
C. 27
D. 36
E. 45

Firstly, observe that  a > b  implies that  ab > ba .
 
 ab  dividing by  ba  yielding the remainder of  r  can be expressed as:  ab = ba*q + r .
 
To obtain the maximum remainder, we should maximize the difference between  ab  and  ba , ensuring that the quotient ( q ) is 1. In this case, the remainder ( r ) would be equal to the difference between  ab  and  ba .
 
For example, if  ab = 98  and  ba = 89 , then the remainder is 9, with  q=1  and  r  being the difference between  ab  and  ba .
 
Similarly, if  ab = 97  and  ba = 79 , the remainder is 18, where  q=1  and  r  is the difference between  ab  and  ba .
 
When  ab = 96  and  ba = 69 , the remainder is 27, again with  q=1  and  r  as the difference between  ab  and  ba .
 
If  ab = 95  and  ba = 59 , the remainder is 36, where  q=1  and  r  is the difference between  ab  and  ba .
 
If  ab = 94  and  ba = 49 , the remainder is 45, with  q=1  and  r  being the difference between  ab  and  ba .
 
However, if  ab = 93  and  ba = 39 , the remainder is 15. Here,  q=2 , hence  r  is not the difference between  ab  and  ba , resulting in a smaller remainder.
 
Thus, the largest possible remainder is 45.

Answer: E

Archit3110 · Answer

given a>b and ab , ba are 2 digit numbers
98 ; 89 ; 9 is remainder
98-79 ; 18 remainder
.
.
94 - 49 ; 45 remainder

option E

gmatophobia · Answer

We can use options to our advantage to solve the question. However, we need to keep in mind that the remainder is always less than the the divisor and is always non negative.

We know in the number ab; a > b

So let's start by taking 98 first. If 98 is the dividend, 89 is the divisor, then the remainder will be 9. However we need the maximum remainder.

This can happen when we maximize the gap between a and b.

Next, we can take ab = 95, ba = 59. In this case the remainder will be 36.

Let's take the numbers farther apart and see if we can maximize the remainder.

ab = 94, ba = 49. In this case the remainder = 45

This is the maximum value in the option available.

Option E.

gmatophobia · Answer

Archit3110  - Are these ab & ba pairs ? If so, none of the pairs are correctly identified.

For example in the "16 ; 61" pair if the divisor is 16, how can the remainder be 45 ?

Archit3110 · Answer

i missed on the part a>b

gmatophobia

Adarsh_24 · Answer

Is this question max possible from given options?

Because I took
ab=ba*q+r
q any positive value

Got 
r=10a+b-10bq-aq = 10(a-bq)+ (b-aq)

Maximum when q is 1. 
r=10(a-b)+(b-a) = 9(a-b)
a-b can be 8 maximum.
So I was thinking 9*8=72

pierjoejoe · Answer

is there anyone who can confirm my solution?
a>b
10a+b = (10b + a) * k + r

let's set
x = 10b + a

than from the first equation

9a+a + 10b -9b = (10b + a) * k + r
(10b + a) + 9b + 9a = (10b + a) * k + r
x + 9a + 9b = x * k + r

9(a+b) = x k +r -x
9(a+b) = x(k-1) + r

meaning that

x = [9(a+b) - r]/(k-1)

now given that x, (a+b), r and k-1 must be all positive integers
to maximize the term on the right of the last equation we have to set
k = 2
r = 0

so the equation becomes
x = 9(a+b)

a+b --> between some values from 0 to 11 (because for 12 x is no more a 2 digit number as asked by the problem)

for every a+b greater than 5 --> x = 10b + a is such that a<b, so we can exclude them
example
x = 9*6 = 54 = 10b + a = 10*5 + 4 --> a = 4, b = 5 --> a<b
x = 9*9 = 81 = 10b + a = 10 * 8 + 1 --> a = 1, b = 8 --> a<b

the greatest multiple of 9 that has 2 digits, such that a>b is 45
x = 45 --> 10 *b + a = 10*4 + 5 --> 5>4 --> a>b

DhanyaAbhirami · Answer

Adarsh_24   adityakaregamba 
Though we are maximising   9(a-b) = r   , we must remember that r is the remainder when divided by   10b+a   , so   r < 10b+a   
When a = 9 and b = 1,   10b+a = 19  .   9(a-b) = 9*8=72   . Notice   9(a-b)   is not less then   10b+a   
When a = 9 and b = 2,   10b+a = 29   .   9(a-b) = 9*7=63   . Notice   9(a-b)   is not less then   10b+a   
When a = 9 and b = 3,   10b+a = 39   .   9(a-b) = 9*6=54   . Notice   9(a-b)   is not less then   10b+a   
When a = 9 and b = 4,   10b+a = 49   .   9(a-b) = 9*5=45   . Notice   9(a-b) < 10b+a   
Hence we choose   r = 45   , Making it the largest remainder possible

bumpbot · Answer

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If r is the remainder when ab is divided by ba, where both ab and ba

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