Find the least number which,when increased by 3,is divisible by 55,88,110.
Answers
Answer: To find the least number which when increased by 3 is divisible by 36, 40 and 64, we first have to find the L.C.M. of 36, 40 and 64
Prime Factorization of numbers
36 = 2*2*3*3
40 = 2*2*2*5
64 = 2*2*2*2*2*2
L.C.M. of 36, 40 and 64 = 2*2*2*2*2*2*3*3*5
= 2880
So, 2880 is the number which is already increased by 3.
Therefore, the number we need is 3 less than 2880
So, the required number is 2880 - 3 = 2877
Hence, 2877 is the least number which when increased by 3 is divisible by 36, 40 and 64.
V)The required number is 9680 which is exactly divisible by 55, 88 , 110.
Step-by-step explanation:
To find : The Greatest 4 digit number which is exactly by 55, 88 , 110?
Solution :
The greatest 4-digit number is 9999.
Now, Find the LCM of 55,88,110.
2 | 55 88 110
2 | 55 44 55
2 | 55 22 55
11 | 55 11 55
5 | 5 1 5
| 1 1 1
Divide 9999 by 440 so that we get the remainder,
Remainder is 319.
Now, Subtract the remainder from 99999.
Therefore, The required number is 9680 which is exactly divisible by 55, 88 , 110
Hence, 2877 is the least number which when increased by 3 is divisible by 36, 40 and 64. V)The required number is 9680 which is exactly divisible by 55, 88 , 110.
Step-by-step explanation: