Mettl Questions on Numbers

Let the numbers be x, 2x and 3x.
The HCF in x, 2x and 3x is x because 1, 2, 3 are prime.
Hence,
x = 12; then the other numbers are 24 and 36.

Since 84 is multiple of 12 hence,any no divided by 84 must be divided by 12..
but, numbers<84 can\'t be divided by 84, hence we divide it by 12 to see what comes as remainder..So..
57
57 / 12 = 9 as remainder...

 

(2^3)^123456=2^370368
Take unit position of power
2^8=256
so unit position is 6

We have to find the Least number, therefore we
find out the LCM of 8, 12, 16 and 20.
8 = 2*2*2;
12 = 2*2*3;
16 = 2*2*2*2;
20 = 2*2*5;
LCM = 2*2*2*2*3*5 = 240;
This is the least number which is exactly divisible by
8, 12, 16 and 20.
Thus,
required number which leaves remainder 5 is,
240+5 = 245.

100! has 23 in it.
Since 100! = 1x2x3.......22x23x24..........99x100.

Thus, when we divide it by 23, the remainder shall be 0.

496/6, We can write it in this form
(6-2)96/6
Now, Remainder will depend only the powers of -2. So,
(-2)96/6,It is same as
([-2]4)24/6, it is same as
(16)24/6
Now,
(16*16*16*16..... 24 times)/6
On dividing individually 16 we always get a remainder 4.
So,
(4*4*4*4............ 24 times)/6.
Hence, Required Remainder = 4.

Let the numbers be 17x and 17y where x and y are co-prime.
LCM = 17xy.
Now, 17xy = 714;
Or, xy = 42 = 6*7;
→x = 6 and y = 7;
Or, x = 7 and y = 6;
1st number = 17*6 = 102;
2nd number = 17*7 = 221;
Sum = 102+119 = 221.

that two digit no is =18
x+y=9
9(10x+y)=2(10y+x)
on solving eqns
x=1,y=8.. no is 18

They will ring together after,
LCM of 48 and 50 secs.
48 = 2*2*2*2*3;
50 = 2*5*5;
LCM = 2*2*2*2*3*5*5 = 1200 secs
= 20 min.
They will beep together at 12:20.

In such case, we first need to find the HCF of 44, 66, 110.
44 = 2*2*11;
66 = 2*3*11;
110 = 2*5*11;
HCF = 2*11 = 22.
Then, the required numbers of rows,
= {(44/22)+(66/22)+(110/22)} = 10.