Question

find the smallest no. which leaves remainder 8and 12 when dibided by 28 and 32

Answer 1

Hii Friend,

28-8 = 20


32-12 = 20

Prime factorisation of 28 = 2 × 2 × 7

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

LCM of 28 and 32 = 224


required Number = LCM of (28,32)-20 => 224-20 = 204

HOPE IT WILL HELP YOU...... :-)

Answer 2

HEYA.....✌✌✌✌✌✌

HERE IS YOUR ANSWER.............

LCM OF 28 AND 32
LCM ====>>224

ADDITION OF REMAINDERS=(8+12)
==========================>20

SO
x+20=LCM
x+20=224
x=224-20
x=204

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HOPE MY ANSWER WOULD HELP YOU

REGARD MADHU@