the least number which is a multiple of 31 and when divided by 15,24,and 32 leaves the remainder 2,11 and19.
Answers
Answer:
(1)Take L. C. M of given no 18,21 and 24
2- use the formula
L.C.M×M−R=23×n
where R is
no. of term
sum of reminders
and M,n are variable
3- use the hit and trial method by putting the values M and check whether the no is divided by 23 or not
we have to find the LCM of the divisors i.e.18,21 and 24.
The LCM of 18, 21 and 24 is 504
and
if we subtract 11 from it, we get 493 and this is the required number that leaves the remainders 7, 10 and 13 with the divisors 18, 21 and 24 respectively.
Now, we have 2 variables and there’s only 1 equation, so obviously we are going to have infinite solutions and we will have to proceed with the trial and error method to obtain the first solution.
If we try with M as 1, we get a number that leaves 10 as the remainder with 23. If we try with M as 2, we get a number that leaves 8 as the remainder with 23. So we expect 0 as the remainder when M=6.
If we try with M as 6, indeed we get 0 as the remainder with the number 504×6–11=3013
Hence required number is 3013.
Step-by-step explanation:
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2387 is the smallest number that is a multiple of 31 that leaves 2, 11, and 19 in its place after being divided by 15, 24, and 32.
Given that,
We have to find what is the smallest number that is a multiple of 31 that leaves 2, 11, and 19 in its place after being divided by 15, 24, and 32.
We know that,
The difference of the numbers we will get the same number
15-2 =13
24-11 = 13
32-19 = 13
The LCM of the number is
15, 24, 32
= 480
Let the required number = 480k - 13
We need to find the multiples of 31
= 480k + (15k-13)
Check for the value of k for which 15k-13 is divisible by 31.
Now, 15×5 = 75-13 = 62 which is divisible by 31.
So, k=5
So, The required number is 480k - 13
= 480(15)-13
= 2387
Therefore, 2387 is the smallest number that is a multiple of 31.
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