Now we have given the number 2000.
Dividing it by 19, we get:
2000 = 105 x 19 + 5
Then:
19 - 5
14
So if we 14 added in 2000, then 2014 will be exactly divided by 19.
So the least number that must be added will be 14.
Let the number be x
1/4 of the number = 1/4(x) =x/4
1/3 of the number = 1/3(x) =x/3
1/4 of a number subtracted from 1/3 of the number gives 12.
x/3-x/4 =12
x/3×4/4 - x/4×3/3 =12
4x/12 - 3x/12 =12
(4x-3x)/12=12
4x-3x=12×12
x=144
Therefore,the number is 144.
Let the number be y
Three fifth of one fourth of a number is 90.
⇒ 3/5 x 1/4 x y = 90
Evaluate LHS:
⇒3/20 y = 90
Multiply both sides by 20:
⇒ 3y = 1800
Divide both sides by 3:
⇒ y = 600
The number is 600.
The least number divisible by 6, 8 and 15
is their L.C.M. which is 120
Now 120 = 2x2x2x3x5
To make it a perfect square, it must be
multiplied by 2x3x5
Required Number=120x2x3x5=3600
separately by 15, 20, 36 and 48 leaves 3 as remainder
in each case.
Required number
= L.C.M. of (15,20,36 and 48) +3
= 720 + 3 = 723
The smallest 3-digit number is 100, which is divisible by 2.
$\therefore$ 100 is not a prime number.
$ \sqrt{101} $ < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11.
$\therefore$101 is a prime number.
Hence 101 is the smallest 3-digit prime number.
(4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11.
(2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11.
(4 + 6 + 1) - (2 + 5 + 3) = 1, not divisible by 11.
(4 + 6 + 1) - (2 + 5 + 4) = 0, So, 415624 is divisible by 11.
19657+33994=53651
let $x$ - 53651 = 9999
Then, $x$ = 9999 + 53651 = 63650
Given Exp. =$ \dfrac{(a^2 + b^2 - ab)}{(a^3 + b^3)} = \dfrac{1}{(a + b)} = \dfrac{1}{(753 + 247)} = \dfrac{1}{1000} $
$ x $ + 3699 + 1985 - 2047 = 31111
$\Rightarrow$ $ x $ + 3699 + 1985 = 31111 + 2047
$\Rightarrow$ $ x $ + 5684 = 33158
$\Rightarrow$ $ x $ = 33158 - 5684 = 27474.
107 x 107 + 93 x 93= (107)2 + (93)2
= (100 + 7)2 + (100 - 7)2
= 2 x [(100)2 + 72] [Ref: (a + b)2 + (a - b)2 = 2(a2 + b2)]
= 20098
2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276
Marking $\left(/\right)$ those which are are divisible by 3 by not by 9 and the others by $\left(X\right) $, by taking the sum of digits, we get:s
2133 $\rightarrow$ 9 $\left(X\right) $
2343 $\rightarrow$ 12 $\left(/\right)$
3474 $\rightarrow$ 18 $\left(X\right) $
4131 $\rightarrow$ 9 $\left(X\right) $
5286 $\rightarrow$ 21 $\left(/\right)$
5340 $\rightarrow$ 12 $\left(/\right)$
6336 $\rightarrow$ 18 $\left(X\right) $
7347 $\rightarrow$ 21 $\left(/\right)$
8115 $\rightarrow$ 15$\left(/\right)$
9276 $\rightarrow$ 24 $\left(/\right)$
Required number of numbers = 6.
Given Exp. = $ \dfrac{(a + b)^2 + (a - b)^2}{(a^2 + b^2)}$ = $\dfrac{2(a^2 + b^2)}{(a^2 + b^2)} $= 2
7429 Let 8597 - x = 3071
-4358 Then, x = 8597 - 3071
------- = 5526
3071
-------
12345679 x 72= 12345679 x (70 +2)
= 12345679 x 70 + 12345679 x 2
= 864197530 + 24691358
= 888888888
When $ n $ is odd, ( x $ n $ + $ a $$ n $) is always divisible by $ x $ + $ a $.
$\therefore$Each one of 4743 + 4343 and 4747 + 4347 is divisible by 47 + 43.
Given Exp.= -84 x $\left(30 - 1\right)$ + 365= -$\left(84 \times 30\right)$ + 84 + 365= -2520 + 449= -2071
Divisor = 5 x 46 = 230
$\therefore$ 10 x Quotient = 230 $\Rightarrow$ =$ \dfrac{230}{10} $= 23
Dividend = [Divisor x Quotient] + Remainder
= [230 x 23] + 46
= 5290 + 46
= 5336.
4500 x $ x $ = 3375 $\Rightarrow$ $ x$ =$\dfrac{3375}{4500}$=$ \dfrac{75}{100} $=$ \dfrac{3}{4} $