Dell Number System, HCF & LCM Questions

sum from 1 to 33 =561
i.e. (n(n+1)/2)
sum from 1 to 8= 36
561-36= 525
525/9 remainder is 3

(9+10+11+12.......33)-(1+2+3...8)
so here
n(n+1)/2-n(n+1)/2
(33x34)/2-(8x9)/2
33x17-4x9
561-36=525
now 525/9=3

Express 3240 as a factors of prime nos

3240 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5


a,b,c in : [3+(3-1)]C (3-1) ways
i.e 5C2 = 10 ways

Nos of ways in which 4 nos of 3's can be distributed
to 3 integers : [4+(3-1)]C (3-1) ways
i.e 6C2 = 15 ways

Nos of ways in which 1 nos of 5 can be distributed to
3 integers : [1+(3-1)]C (3-1) ways
i.e 3C2 = 3 ways

So total ways = 10 x 15 x 3 = 450

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

Let the numbers be 37a and 37b.
Then , 37a x 37b =4107 => ab = 3.
Now co-primes with product 3 are (1,3)
So, the required numbers are (37 x 1, 37 x 3)
i.e, (37,111).
Therefore, Greater number = 111.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4,
which is multiple of 7.

Least value of k for which (90k + 4)
is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4 = 364.

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7
5.25 meters=525 cm = 5 × 5 × 3 × 7
Hence common factors are 3 and 7
Hence LCM = 3 × 7 = 21
Hence largest size of square tiles that can
be paved exactly with square tiles is 21 cm.

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40)
and (5 x 40).

Hence, required H.C.F. = 40.

Required number = H.C.F. of (1657 - 6)
and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

LCM of 36,40,48 is 720
720/60=12

99(100 - 1) = 9900-99= 9801
9801(100 - 1) = 980100-9801= 971299
971299(100 - 1) = 97129900 - 971299 = 96157601
.....
.....
observe the pattern, 98, 97, 96, ....
for power of 2, 3, 4, ...
So for 90 the power could be 10.
For 11, we get a number starts with 8

as there are only 3 squares b/w 2006-2300
that are 45^2=2205, 46^2=2116,
47^2=2209
now this series is an A.P. with 7 differnce.
so the difference of these squares & 2006
must be a multiple of 7.
only 2209-2006=203
which is multiple of 7.
so correct answer is 1.

There is a standard way of solving these questions quickly, which you should mug up/write in notes:

dividend = divisor × quotient + remainder

• The remainder when (m + n) is divided by 12 is 8


• What will be the minimum possible value of (m+n) when the quotient is 1

• (m + n) = 12*1 + 8


• (m + n) = 20

• (m - n) = 12*1 + 6


• (m - n) = 18

• 2m = 38


• m = 19


• replacing m in any of the above equation we will get n = 1

Let three consecutive odd integers be
x,x+2,x+4
4x=6+2(x+4)
x=7
x+4=11

Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 36
9(x - y) = 36
x - y = 4.

1/3 of 1/4 of x = 15
x = 15 x 12 = 180.

So, required number = (3/10)x 180
= 54.

If we consider the third term to be ‘x”
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d).
Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
x + 9d will be the 12th term.