A number which is divisible by both 3 and 4 is divisible by 12 . Hence (a).

How many factors are there in 432?

Prime factors of the number 432 = (2 ^{4} *3 ^{3} )

Number of factors = (4+ 1)* (3+1) 20. Hence (b).

The remainder is 15 . The remainder is one third of quotient, then quotient = 45.The divisor = 2 * 45 = 90. Number = (90 * 45) + 15 = 4065. Hence (d).

LCM and HCF of two numbers are 396 and 12 respectively. If one of the numbers is 36, find the other number.

LCM * HCF = product of two numbers .

396 * 12 = 36 * x. x= 132. H ence (c).

what is the unit digit of 1293^{323} + 2095^{496}

Unit digit of 1293 ^{323} = 7

Unit digit of 2095 ^{496 } = 5

Unit digit of 1293 ^{323} + 2095 ^{496} = 7 + 5 = 12 => 2. Hence (b).

Multiply 27 by 2 then add 16 to it and divide it by 10 and add it with 7. What is the result?

27 * 2 = 54 + 16 = 70

70/10 = 7 + 7 = 14 . Hence (c).

LCM of 5, 4, 3 and 9 = 180. The nearest multiple is 3780. Therefore, 34 should be added.

Hence (b).

Hence (b).

What is the remainder of (21^{171} + 11^{171}) /32

propertyà Remainder of (a^{n} + b^{n}) / (a + b) = 0 where n is odd. Hence (c).

How many 3’s are used for numbering the pages in a article having 400 pages?

Number of possibilities of 3’s in unit digit is 40 times

Number of possibilities of 3’s in ten’s digit is 40 times

Number of possibilities of 3’s in Hundred’s digit is 100 times,

So, the total numbers is 180. Hence (d).

What is the remainder when 4^{3291} is divisible by 6?

Remainder of 4 power any number divided by 6 = 4.

R( 4/6) = 4, R(16/6) = 4. Hence (a).

Convert decimal to octal 920_{10}

( 920 ) _{10 } _{=} ( 1630) _{8} Hence (d).

HCF of two numbers is 18 and sum of the numbers is 216. How many such pairs are possible?

Let the two numbers be x and y.

X = HCF * a = 18a

Y = HCF * b = 18b

X + y = 216 = 18(a+b) => a+b = 12

Then, 2 such pairs are possible (1,11) and (5,7)

For example the pair is (1,11), then x=18*1 =18 and y = 18*11 = 198, x+y = 18 + 198 à 216.Hence (a).

From options first number is 11.check out the unit digits of products of the last three and first three. Hence (a).

If \(\)\({(a+b)^{n+1} \over {a^{n^{2}}+b^{n^{2}}}}\) is the arithmetic mean of a and b, then the value of n is

When n=0, the equation becomes \({a+b \over {2}}\) Which is the arithmetic mean of a and b. Hence (b).

When a number is successively divided by 3, 4,9, it leaves a remainder 2, 1 and 7 respectively. Find the smallest such number.

If x is the number and q _{1} , q _{2} , q _{3 } are the successive quotients, then x =3q _{1} + 2

q _{1} = 4q _{2} + 1, q _{2} = 9q _{3} + 7. The smallest number will be if q _{3} = 0, x = 89. Hence (d).

The sides of a quadrilateral field are 2783 metres, 3841 metres, 3257 metres and 2668 metres respectively. Find the length of the greatest tape which can measure all the four sides.

Great est length of tape which can measure all four fields will be HCF of four numbers.

The HCF is 23 . Hence (a).

Find the unit digit of (38^{12})

(38 ^{12} ) = 38 ^{132 } The cyclicity of 8 is 4.Remainder when divi ded power by 4 = 0.Unit digit =6 . Hence (c).

The sum of the digits of a two digit number is 9.If the digits are reversed, the reversed number is \({2 \over {9}}\) of the original number. Find the two- third of original number.

Sum of the original number = 9 -> 10x + y

Reversed number = 10y + x = \({2 \over {9}}(10x + y)\)

\({x \over {y}}\) = \({8 \over {1}}\) , original number = 10x + y = 81.Two third of number = 54. Hence (b).

Find the last two digits of 527^{325}

527^{325} = 527 * 527^{324}

= 527 * (527^{4})^{81}

= 527 * (…41)^{81}

= 527 * (….41)

= (….07) Hence (d).

Find the last two digits of 91^{712}

The unit digit of the given number is 1.

But we have to find the ten’s digit of that number.

For that we have multiply the ten’s digit of the given number by the unit digit of its power.

that is 9 * 2 = 1 **8** . So the l ast two digit of the number is 8 1. Hence (a).

Find the product of the factors of 520.

Number of factors of 520 = 2 ^{3 } * 5 * 13 = (3+1) * (1+ 1)* (1+1) = 16 factors

Product of the factors = N ^{(} ^{no of factors /2)} = 520 ^{(16/2)}

--> 520 ^{8} Hence (d).

Find the sum of the factors of 2842.

2842 = 2 * 7 ^{2} * 29

Sum of the factors = (\({2^{1+1}- 1 \over {2-1}}\) )( \({7^{2+1}- 1 \over {7-1}}\) )( \({29^{1+1}- 1 \over {29-1}}\) ) = 3*57*30 = 5130 . Hence (c).

According to wilson's property R( \({(p-1)! \over {p}} )= p-1\) , where p is a prime number. Hence (a).

Unit digit of 97 ^{122} = 9, then R(9/10) = 9 Hence (c).

Find the remainder of 37^{42} when divided by 43.

According to Fermat's theorem,

R( \({a^{p-1} \over {p}}\) ) = 1 , where a,p are co-prime numbers. then R( \({37^{43-1} \over {43}}\) ) = 1 Hence (a).