## IBM HCF Practice Problems

Here are four problems which are conceptually similar but are slightly different in nature from each other.

Question 1

Find the largest number that will divide 3532, 5382, 5937 and gives the same remainder.

a)185 b)815 c)435 d)585

Solution:

To solve such problems, we can adopt the following method.
Let x1,x2 and x3 be the given numbers put in ascending order and we have to find the largest number that will divide x1,x2 and x3, leaving the same remainder.
Then our answer would be hcf of [(x2-x1),(x3-x2),(x3-x1)]

the required largest number = h.c.f [(5937-3532),(5382-3532),(5937-5382)]
= h.c.f (2405,1850,555)
= 185.

Question 2

Find the greatest number that exactly divides 7667, 4603, 12263 when each of these is reduced by 7.

a) 2298 b) 1572 c) 1532 d) 2546

Solution:

(Actually this is a simple problem dealing with finding just HCF of the numbers after subtracting 7 from each of these numbers. Since it immediately follows the I problem, some readers could think it as complex as the first.)

the required number = h.c.f of [(4603-7),(7667-7),(12263-7)] (given that the numbers leaves 7)
= h.c.f of (4596, 7660, 12256)
= 1532.

Question 3

Consider three numbers 2331,1906 and 1806. Find the ratio between the largest number that divides each of the three numbers giving a same remainder R to the smallest number that gets divided by the three numbers when R is subtracted from each of them.

a) 2:21204 b) 1:42408 c) 2:42408 d) 1:21204

Answer : b) 1 : 42408

Solution :

I Part : Finding the largest number that divides each of the three numbers giving a same remainder :

This part is similar to the first problem.

Arranging the numbers in ascending order, we get, x1 = 1806 , x2 = 1906 and x3 = 2331

The required largest number = hcf of [(x2-x1),(x3-x2),(x3-x1)] = hcf of [(1906 - 1806),(2331 - 1906 ),(2331 - 1806)]
= hcf of [100, 425, 525]
= 25 ...(1)

Now we know that when each of 2331,1906 and 1806 will give same remainder R when divided by 25

To find R, let us divide any of the numbers, say 1906 by 25. The remainder we get would be 6. Therefore R = 6.

II Part : Finding the smallest number that gets divided exactly by each of the three numbers when R is subtracted from each of them :

Subtracting R from each of the numbers 2331,1906 and 1806, we get, 2325,1900 and 1806

Smallest number that gets divided exactly by 2325,1900 and 1806 will be the LCM of [2325,1900 and 1800] = 1060200 ...(2)

Ratio between (1) and (2) = 25 : 1060200 = 1 : 42408

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