Aptitude - Problems on H.C.F and L.C.M - Discussion

4 * 1304 + 1 = 1305
4 * 4664 + 1 = 4665
4 * 6904 + 1 = 6905.

Very good and helpful. Thanks all.

How would we know that H.C.F. of 3360, 2240 and 5600 = 1120.

First do this(4665 - 1305) , (6905 - 4665) and (6905 - 1305). we ll get 3360,2240,5600. then divide by 10.. now
336=2*2*2*2*3*7*10
224=2*2*2*2*2*7*10
560=2*2*2*2*5*7*10

So take the common multiples.. now we got 2*2*2*2*7*10=1120( this 1120 is "N")

So, question asked here is sum of digits "N"??
"N" is 1120; sum of digits "N" is 1+1+2+0=4.
Tats it!;)

Let N be the greatest number that will divide 17, 41 and 77, leaving the same remainder in each case. Then the sum of the digits in N is:

It is easy to understand this with a simple example, and the logic behind this is so obvious.
Take number 12 (you can get any number you desire and practice it!).

Dividend = Divisor X Quotient + remainder (when the remainder is a constant!).

12 X 1 + 5 = 17.
12 X 3 + 5 = 41.
12 X 6 + 5 = 77.

Now, take the differences of 17, 41 and 77 in order considering the smallest, the middle and the largest numbers.

(41-17) = 24; (77-17) = 60; (77-41) =36.

Now take 24, 60, and 36 and find the HCF or HCD of these numbers: it is equal to 12.
Look at the "Divisor" used at the beginning. So, the divisor of that arrangement is equal to the HCF of 24, 60 and 36. When the remainder is constant, we can observe this pattern with any set of numbers arranged in this way.

The sum of digits of N = 1+2 =3.

Thank you for explaining the answer @Greeshma.

The easiest way to find this answer is subtracting until you get repeated value.

(4465-1305 = 3360.
6905-4665 = 2240.
6905-1305 = 5600).

(3360-2240 = 1120.
5600-2240 = 3360.
5600-3360 = 2240).

(2240-1120 = 1120.
3360-1120 = 2240.
3360-2240 = 1120).

So, the ans is 1120.
1+1+2+0 = 4.

1305, 4665, 6905 -----> These are the numbers.

In this question why we subtract the numbers because here, it is said that they have the same remainder.

So, 1305 = hcf * x + remainder -----> (1),

4665 = hcf * y + remainder -----> (2),

6905 = hcf * z + remainder -----> (3),

Where x, y, z are respective quotients.
Now we take (2) - (1).
3360 =hcf (y-x) -----> (4),
From (3) - (2) we have
2240 = hcf (z - y) -----> (5),
From(3) - (1) we have
5600 = hcf (z - x) -----> (6),
Now we can see from the equation (4), (5), (6) hcf of the three numbers is the hcf of the three numbers. So the hcf of 6905, 4665 and 1305 = hcf of the three numbers 5600, 3360, 2240 = 1120.

Any no. Can be represented as of the form D*q+r right. so we want remainder same in each case. So
1305=N*q1+r
4665=N*q2+r
6905=N*q3+r
Now, now subtracting 2 eq from 1 then ..3 to 2 then 3 to 1 we get N*(q2-q1)=3360.
Similarly, we will get the rest ones. now here, we just want the greatest no. Which divides all this num3360 2240 nd 5600 ....so simply take the hcf off all. We will get N=1120.

I hope this helps you.

8*163 = 1304 thus 1 remainder
583*8 = 4664 thus 1 remainder
863*8 = 6904 thus 1 remainder
So the answer should be 8.