Let the number be 27a and 27b
Then, 27a + 27b = 216
or a + b = 216/27 = 8
? Values of co-primes with sum 8 are (1, 7) and (3, 5)
So, the number are (27 x 1 , 27 x 7) i.e. (27, 189)
Required length = H.C.F of 700, 385, 1295 cm
= 35 cm.
Find the H.C.F. of 84 and 180, which is 12 and this is the required answer .
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.
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 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.
Let the number be 12a and 12b
Then 12a x 12b = 2160
or ab = 15
? Values of co-primes a and b are (1, 15), (3, 5)
So, the two digit number are 12 x 3 and 12 x 5 i.e. 36 and 60 .
Product of two numbers = LCM of two numbers x HCF of two numbers.
? N x 32 = 160 x 16
? N = 80
Product of two numbers = LCM of two numbers x HCF of two numbers.
? 1600 = LCM x 5
? LCM = 320
H.C.F. of two numbers divides their L.C.M. exactly. Since 16 is not a factor of 136, it states that there does not exist any pair of numbers with H,C.F. 16 and L.C.M. 136.
Here (16 - 3) = 13,
(18 - 5) = 13 and
(21 - 8 ) = 13
So, required number = (L . C . M of 16, 18, 21) - 13
= (1008 - 13)
= 995
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