find the smallest number when divided by 161 207 and 184 leaves remainder 21 in each case
Answers
Here, I am finding LCM by prime factorization method:
Write 161 , 207 and 184 into product of primes
161 = 7 × 23 = 7¹ × 23¹
207 = 3 × 3 × 23 = 3² × 23¹
184 = 2 × 2 × 2 × 23 = 2³ × 23¹
LCM of ( 161 , 207 , 184 ) =23 × 7 × 3² × 2³ = 23 × 7 × 3 ×3 × 2 × 2 ×2 = 11592
LCM of ( 161 , 207 , 184 ) = 11592
Required smallest number which when divided by 161 , 207 & 184 leaves remainder 21 = 11592 + 21 = 11613
Hence, The required smallest number which when divided by 161, 207 and 184, leaves a remainder of 21 in each case is 11613.
Verification :
11613 / 161
Quotient = 72 and Remainder = 21
11613 / 207
Quotient = 56 and Remainder = 21
11613 / 184
Quotient = 63 and Remainder = 21
HOPE THIS ANSWER WILL HELP YOU...
Answer:
Step-by-step explanation:
To find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in all case, we have to compute the L.C.M. of 161, 207 and 184 and then we will add 21 in the L.C.M. of 161, 207 and 184. The number we have after adding will be the required number.
2 | 161, 207, 184
|. ______________
2 | 161, 207, 92
|______________
2 | 161, 207, 46
|______________
3 | 161, 207, 23
|______________
3 | 161, 69, 23
|______________
7 | 161, 23, 23
|______________
7 | 23, 23, 23
|______________
23 | 1, 1, 1
|
L.C.M. of 161, 207 and 184 is 2*2*2*3*3*7*23
= 11592
So. L.C.M. of 161, 207 and 184 is 11592
Now, we will add 21 and 11592
11592 + 21 = 11613
The required smallest number is 11613, which when divided by 161, 207 and 184, leaves a remainder of 21 in each case.
Let us check our answer.
11613 ÷ 161
Quotient = 72 and Remainder = 21
11613 ÷ 207
Quotient = 56 and Remainder = 21
11613 ÷ 184
Quotient = 63 and Remainder = 21
Hope it helps