Question

find the smallest number which when divided by 1088, 640 and 100 leaves 3 as reminder in each case in

Answer 1

To Find :-

• The smallest number which when divided by 1088, 640 and 100 leaves 3 as reminder in each case ?

Solution :-

we know that,

• The smallest number which when divided by a, b and c leaves r as reminder in each case is = LCM(a,b and c) + r .

so,

Prime factors of 1088 , 640 and 100 are :-

→ 1088 = 2⁶ * 17

→ 640 = 2⁷ * 5

→ 100 = 2² * 5²

LCM = 2⁷ * 5² * 17 = 128 * 25 * 17 = 54400 .

therefore,

→ Required number = 54400 + 3 = 54403 (Ans.)

Learn more :-

वह छोटी से छोटी संख्या बताईये जिसमे 7,9,11 से भाग देने पर 1,2,3 शेष बचे

https://brainly.in/question/9090122

Answer 2

Given:

find the smallest number which when divided by 1088, 640 and 100 leaves 3 as a reminder in each case.

To find:

The required number

Solution:

Prime factors of 1088:

1088 = 2x2x2x2x2x2x17 = 2⁶ x 17

Prime factors of 640:

640 = 2x2x2x2x2x2x2x5 = 2⁷ x 5

Prime factors of 100:

100 = 2x2x5x5 = 2² x 5²

LCM = 2⁷ x 5² x 17 = 128 x 25 x 17 = 54400

54400 + 3 = 54403

Hence, the required answer is 54403