Given, a = 12, b =16 , c =18, k = 5
According to the formula,
Required number =(LCM of a, b and c) + k
= (LCM Of 12, 16, 18) + 5
LCM of 12, 16, 18 is
? LCM = 2 x 2 x 3 x 4 x 3 = 144
? Required number =144 + 5 = 149
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37.
? Number to be added = 60 - 37 = 23
Given that, product of two numbers =1500
HCF = 10
According to the formula,
Product of two numbers = HCF x LCM
? 1500 = 10 x LCM
? LCM = 1500/10 =150
Let numbers are 2N and 3N.
According to the question, 6N = 48
? N = 8 ( ? LCM = 6N )
? Required sum = ( 2N + 3N ) = 5N
= 5 x 8 = 40
LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 x 13, then the same number must be exactly divisible by their LCM or by (11 x 13).
Let numbers are 5N and 6N.
Now, HCF of these two numbers is N.
We know that,
LCM x HCF = Product of two numbers
? 480 x N = 5N x 6N
? 480N = 30N2
? N =16
The number when divided by 9, 11, 13 leaving remainder 6 = (i.c.m. of 9, 11, 13) + 6 = 1293
Required least number = 1294 - 1293 = 1.
The least common multiple of 38 and 57 is 114 and the multiple which is between 900 and 1000 is 912.
Now, 912 + 23 i,e 935 lies between 900 and 1000 and when divided by 38 and 57 leaves in each case 23 as the remainder.
Therefore, 935 is number required.
From the question we see that the second number is a common factor of the two products and since the numbers are prime to one another. It is their . H. C. F . and is, therefore, 19.
? The first number = 437 / 19 = 23
and the third number = 551 / 19 = 29
Hence, the number are 23, 19 and 29
? Sum = 23 + 19 + 29
= 71.
The L.C.M of 9, 10, 15 = 90
On dividing 1936 by 90, the remainder = 46
But a part of this remainder = 7
Hence, the two numbers = 46 - 7 = 39.
Find the H.C.F. of 84 and 180, which is 12 and this is the required answer .
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