Question

What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

Answer 1

Step-by-step explanation:

Let the no. be x

then, the no. to be divided=1936-x

now, given,

(1936-x) leaves a remainder 7 when divided by 9,10 and 15

therefore, if we subtract 7 from (1936-x),

the number, (1936-7-x) will be perfectly divisible by 9, 10, and 15.

Hence,

(1929-x)÷9=(1929-x)÷10=

=(1929-x)÷15

(1929-x)÷9=(1929-x)÷10

=19290-10x=(1929)9 - 9x

=19290-(1929)9=10x-9x

=1929=x

x=1929

(1929-x)÷10=(1929-x)÷15

=(1929)15-15x=(1929)10 - 10x

=(1929)15-(1929)10=15x-10x

=(1929)5=5x

x=1929

Thus, the least number is 1929

proof,

1936-1929=7

7÷9=7 as remainder

7÷10=7 as remainder

7÷15=7 as remainder

Awakward ans... I know, but it is the correct anwer....

Hope it works