Find the sum of three numbers which are prime to one another such that the product of the first two is 437 and that of the last two is 551?

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.

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.

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

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 .

Required length = H.C.F of 700, 385, 1295 cm
= 35 cm.

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)

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 .