Question

Find the smallest number which when divided by 57, 76 and 190 leaves the remainder 1 in each case

Answer 1

Answer:

The smallest number is 1141

Step-by-step explanation:

To find the smallest number which is divisble by 57, 76 and 190, we have to find LCM of 57, 76 and 190.

LCM of 57, 76 and 190 = 2 x 2 x 3 x 5 x 19 = 1140

Now since the number leaves remainder 1 in each case, we have to add 1 to get the required smallest number.

So, 1140 + 1 = 1141

Hope This Helps You!

Answer 2

Answer:

57, 76 and 190 leave a remainder 1, so the numbers are 57-1=56, 76-1=75 and 190-1=189

Now we find the HCF of 75 and 189

By Euclid's division algorithm, we have

189=75*2+39

75=39*1+36

39=36*1+3

36=3*12+0

Therefore, HCF(75&189)=3

Now we find the HCF of 3 and 57

By Euclid's division algorithm, we have

57=3*19+0

Therefore, HCF (3 and 57) = 3

Therefore, the required smallest number is 3

HOPE IT HELPS YOU!