Question

Find the smallest number which when divided by 42, 56 and 35 leaves the same remainder 5.

Answer 1

subtract 5 from all 3

42-5=37

56-5=51

35-5=30

finding hcf of 37 and 51 by euclids division lemma

51=37×1+14

37=14×2+9

14=9×1+5

9=5×1+4

5=4×1+1

4=1×4+1

hcf =1

hcf of 1 and 30

30=1×30+0

hcf=1

therefore the smallest no. which divide 42,56,35 leaving the remainder 5 to all is 1

Answer 2

Hello

answer for your question 1

I will send whole solution later