Question

Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32
(Along the answer plz tell the steps to solve this question)

Answer 1

Hey there !!

Here , we are asked to find the smallest number which leaves some remainders when divided by certain numbers.
To find the smallest number , follow these steps :-

1) Subtract the remainders from the given numbers:
28-8 = 20
32 -12 = 20

Both of them give 20

2) Find the LCM of 28 and 32
28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2

LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224


3) Smallest no: which leaves remainder 8 and 12 when divided by 28 and 32 = LCM -20 = 224 - 20 = 204

204 is the answer !!!

Answer 2

Let smallest number is x. Accordingly:-.

x = a×28+8……………(1)

x= b× 32 +12……………(2)

from eq.(1)& (2)

28a+8 = 32b +12

28a-32b= 4

7a-8 b = 1

7a =8b +1

a= (8b+1)/7………………..(3)

If b=1 then a is not whole number.

b=2 “ “ “ “ “ “ “ “ “ “ “ “ ‘ “ “

× × × × × × × × × × × × × × × × ×

b=6 then a =7 . which is also whole number.

On putting a=7 in eq.(1) or b= 6 in eq. (2)

x=28×7 + 8

x= 196 + 8

x = 204