a+1=b

=> ba=1.

and 11b64 is divisible by 11

=> (4+b+1)(6+1)=0

=> b2=0

=> b=2.

so, a=1

=>(a+b)= 3.

Let the smaller number be x.

Then larger number = (x + 1097)

x + 1097 = 10x + 17

9x = 1080

x = 120

sum=n(n+1)/2

sum=55

n^2+n=55*2

n^2+n110=0

(n10)(n+11)=0

n=10,11,neglect negative

wer =10

Required number = (333+222)×3×111+62

= 184877

Formula is n(n+1)/2,

Here n=75.

So the wer is 2850

These numbers are 28, 35, 42,…., 133.

This is in A.P. in which a= 28, d=(3528)= 7 and L=133.

Let the number of there terms be n. then, Tn=133

a+(n1)d=133 by solving this we will get n=16.

133 is divisible by 7.

Rest of numbers is not divisible by any numbers except itself and 1.

Required difference = (8000 80)

= 7920

On dividing 457 by 11, remainder is 6.

Required number is either 451 or 462.

Nearest to 456 is 462.

Divide 2458 by 13 and we get remainder as 1.

Then 131=12.

Adding 12 to 2458 we get 2470 which is divisible by 13.

Thus wer is 1.

Let the numbers be x and y.

Then, xy = 186 and x2 + y2 = 436.

=> (x y)

2 = x2 + y2 2xy

= 436 (

2 x 186)

= 64

=> x y

= SQRT(64)

= 8.

Let the three integers be x, x + 2 and x + 4.

Then, 2(x+2) = (x + 4) + 6

=> x = 6.

Third integer = x + 4 = 10.

5465 is divisible by 5.

So 6K4 must be divisible by 3.

So (6+K+4) must be divisible by 3.

K = 2

(6475/55)

Remainder =40

647540=6435

Required numbers are 18,24,30,.....84

This is an A.P a=18,d=6,l=84

84=a+(n1)d

n=12

4digit

numbers divisible by 7 are: 1001, 1008, 1015….. 9996.

This is an A.P. in which a=1001, d=7, l=9996.

Let the number of terms be n.

Then Tn=999@.'. a+(n1)d=9996

=> 1001+(n1)7= 9996

=>(n1)7=8995

=>(n1)=8995/7= 1285

=> n=1286.

.'. number of terms =1286.

Let 5^32=x.

Then (5^32+1)=(x+1). Let (x+1) be completely divisible by the whole number Y.

then (5^96+1)=[(5^32)^3+1]=>(x^3+1)=(x+1)(x^2x+1) which is completely divisible by Y.

since (x+1) is divisible by Y.

= 96 x 96 + 84 x 84 = (96)2 + (84)2

= (90 + 6)2 + (90 6)2

= 2 x [(90)2 + (6)2]

=16272

6 = 3 x 2.

Clearly, 13 * 4 is divisible by 2.

Replace * by x.

Then, (1 + 3 + x + 4) must be divisible by 3.

So, x = 1.

x = 28955 4017

= 24938.

=> (2/3)*(3/4)*x = 24

=> x=48,1/3x = 16

= (1004)2+(996)2=(1000+4)2+(10004)2

= (1000)2 + (4)2 + 2*1000*4 + (1000)2 + (4)2 2*100*4

= 2000000 +32 = 2000032

(7n2 + 7n) = 7n(n + 1), which is always divisible by 7 and 14 both, since n(n + 1) is always even.

=> (xn 1) will be divisible by (x + 1) only when n is even.

=> (36^11 1)

= {(6^2)^11 1}

= (6^22 1),which is divisible by (6 +1)

i.e., 7.

Let the numbers be x,y.

=> x2+y2=81,

=> 2(x+y)=40,

=> (x+y)2=81+40=121,

=> x+y=sqrt(121)=11

Let the given number be 597xy6.

Then (5+9+7+x+y+6)=(27+x+y) must be divisible by 3

And, (6+x+9)(y+7+5)=(xy+3) must be either 0 or divisible by @xy+3=0

=> y=x+3 27+x+y)

=>(27+x+x+3)

=>(30+2x)

=> x = 3 and y = 6.

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264,396,792 are divisible by 132.

Required wer =3

(x^n+1) is divisible by (x+1), when n is odd.

.'. (55^55+1) is divisible by (55+1)=5@when (55^55+1)+54 is divided by 56, the remainder is 54.

Let P = 5x + 2.

Then 3P = 15x + 6

= 5(3x + 1 ) + 1

Thus, when 3P is divided by 5, the remainder is 1.

let take 2 consecutive even numbers 2 and 4.

=> (4*4*4)(2*2*2)=648=56 which is divisible by 4.

Let the ten's and unit digit be x and 8/x respectively.

Then, 10x + 6/x + 45 = 10 x 6/x + x

=> 10x2 + 6 + 45x = 60 + x2

=> 9x2 + 45x 54

= 0

=> x2 + 5x 6

= 0

=> (x + 6)(x 1)

= 0

=> x = 1

So the number is 16

=> x+y=13, xy=5

Adding these 2x =18

=> x=9, y=4.

Thus the number is 94

3621 x 137 + 3621 x 63 = 3621 x (137 + 63)

= (3621 x 200)

= 724200

suppose that on dividing the given number by 234,

we get quotient=x and remainder= 43

then, number= 234*x+43----->(1).

=> (13*18x)+(13*3)+4

=> 13*(18x+3)+4.

So, the number when divided by 13 gives remainder=4.

=> x+y=30

=> xy=20

=> (x+y)2(xy)2 = 4xy

=> 4xy=302202=500

=> xy=500/4=125