A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible reminder?

Question

Q:- A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible reminder?

Answer 1

The largest possible remainder occurs on 49 vs 94.

 

94 / 49 = 1 with a remainder of 45.

 

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Answer 2

Let's pick a number to wrap our heads around the situation.

 

number = 21

reversed = 12

 

larger/smaller = 21/12

 

21/12 = 1 + 9/12  where 1 is the quotient, 9 is the remainder, 12 is the divisor

 

Knowing how division works, the largest the remainder can be is 11 in this case.

Or one less than the divisor in any case. Knowing this fact, let's start!

 

 

Let's define our two-digit number:

10T + U = value of the number with

T = digit in tens place

U = digit in units place

 

reversed value:

10U + T

 

Since it is not defined, let's say the tens digit is larger than the units digit:

T > U which makes

10T + U  > 10U + T

 

Now, let's divide larger by smaller with Q= quotient and R = remainder:

 

[ 10T + U ] / [ 10U + T ] = Q + R/ [ 10U + T ]  

 

The remainder must be a natural number at least one less than the divisor (as initially stated)

 

R &lt= 10U + T - 1  

 

when T > U

 

The largest the remainder can be is 10U + T - 1 or one less than the smaller number.