Question

A 2 digit number is reversed. the larger of the two numbers is divided by the smaller one. what is the largest possible remainder?

Answer 1

Answer:

Given : A 2 digit number is reversed. the larger of the two numbers is divided by the smaller one.

To find : What is the largest possible remainder?

Solution :

Let's define our two-digit number:

10x + y = value of the number with

x = digit in tens place

y = digit in units place

Reversed value:

10y+ x

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

x > y which makes,

10x + y  > 10y + x

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

$\frac{(10x+y)}{10y+x}=\frac{Q+R}{10y+x}$

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

R= 10y+x - 1  

when x > y

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

For example -

Number = 21

Reversed = 12

$\frac{\text{Larger}}{\text{Smaller}}=\frac{21}{12}$

$\frac{21}{12}=1+\frac{9}{12}$

Where 1 is the quotient, 9 is the remainder, 12 is the divisor.