1. If 146 Is divisible by 5^{n}, and then find the maximum value of n.

a) 34

b) 35

c) 36

d) 37

Explanation:

$$\eqalign{ & = {\frac{{146}}{5}} + {\frac{{146}}{{{5^2}}}} + {\frac{{146}}{{{5^3}}}} \cr & = 29 + 5 + 1 \cr & = 35 \cr} $$

Note: We have taken integral value only, not the fractional. For example $$\frac{{146}}{5}$$ = 29.2 but we have taken 29 and so on.

2. Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took $$\frac{1}{3}$$ of the mints, but returned four because she had a monetary pang of guilt. Fatima then took $$\frac{1}{4}$$ of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

a) 38

b) 31

c) 41

d) None of these

Explanation:

$$\eqalign{ & {\text{Number of mint before Eswari has taken}}, \cr & = \left( {x - {\frac{x}{2}} } \right) + 2 = 17 \cr & x = 30 \cr & {\text{Number of mint before Fatima has taken}}, \cr & = \left( {x - {\frac{x}{4}} } \right) + 3 = 30 \cr & x = 36 \cr & {\text{Number of mint before Sita has taken}}, \cr & = \left( {x - {\frac{x}{3}} } \right) + 4 = 36 \cr & x = 48 \cr & {\text{There}}\,{\text{were}}\,{\text{48}}\,{\text{mints}}\,{\text{originally}}{\text{.}} \cr} $$

3. Some birds settled on the branches of a tree. First, they sat one to a branch and there was one bird too many. Next they sat two to a branch and there was one branch too many. How many branches were there?

a) 3

b) 4

c) 5

d) 6

Explanation: When the birds sat one on a branch, there was one extra bird.

When they sat Two to a branch one branch was extra.

Go through the options one by one;

Checking option (a);

If there were 3 branches, there would be 4 birds.

(This would leave one bird without branch as per the question)

When 4 birds would sit Two to a branch there would be one branch free

(as per the question).

The option 'a' is correct.

4. If we divide the unknown two-digit number by the number consisting of the same digits written in the reverse order, we get 4 as quotient and 3 as remainder. If we divide the required number by sum of its digits, we get 8 as a quotient and 7 as a remainder. Find the number?

a) 81

b) 91

c) 71

d) 72

Explanation: Go through the options one by one,

Checking option (c);

$$\frac{{71}}{{17}}$$ = 4 quotient and remainder 3.

$$\frac{{71}}{8}$$ = 8 quotient and remainder 7 as remainder.

5. The last three-digits of the multiplication 12345 × 54321 will be

a) 865

b) 745

c) 845

d) 945

Explanation: If we multiply the last three digits of each terms we get the last three digits.

345 × 321 = 110745

The last three digits are 745

6. Find the least number which will leaves remainder 5 when divided by 8, 12, 16 and 20

a) 240

b) 245

c) 265

d) 235

Explanation: 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

Required number which leaves remainder 5 is,

240 + 5 = 245

7. 7^{6n}- 6^{6n}, where n is an integer >0, is divisible by

a) 13

b) 127

c) 559

d) All of these

Explanation:

$$\eqalign{ & {7^{6n}} - {6^{6n}} \cr & = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr} $$

Clearly, it is divisible by 127, 13 as well as 559

8. After the division of a number successively by 3, 4 and 7, the remainder obtained is 2, 1 and 4 respectively. What will be remainder if 84 divide the same number?

a) 80

b) 75

c) 42

d) 53

Explanation: As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)

The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}

Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]

On simplifying,

[3 × {4 × (7x + 4) + 1} + 2]

= 84x + 53

We get the final number 53 more than a multiple of 84. Hence, if the number is divided by 84,

The remainder will be 53

9. Find the remainder when 2^{256} is divided by 17.

a) 1

b) 16

c) 14

d) None of these

Explanation:

$$\eqalign{ & \frac{{{2^{256}}}}{{17}} \cr & {\text{We}}\,{\text{can}}\,{\text{write}}\,{\text{it}}\,{\text{as}}:\,{\left( {{2^4}} \right)^{64}} \cr & \frac{{{{16}^{64}}}}{{17}} \cr & {\text{Individually, when 16 is divided by 17,}} \cr & {\text{gives a negative reminder of - 1}}{\text{.}} \cr & {\text{Required Remainder}}, \cr & {\left( { - 1} \right)^{64}} = 1 \cr} $$ .

10. Find the remainder when 4^{96} is divided by 6.

a) 0

b) 2

c) 3

d) 4

Explanation: $$\frac{{{4^{96}}}}{6},$$ we can write it in this form

$$\frac{{{{\left( {6 - 2} \right)}^{96}}}}{6}$$

Now, Remainder will depend only the powers of -2.

$$\frac{{{{\left( { - 2} \right)}^{96}}}}{6},$$ it is same as

$$\frac{{{{\left( {{{\left[ { - 2} \right]}^4}} \right)}^{24}}}}{6},$$ it is same as

$$\frac{{{{\left( {16} \right)}^{24}}}}{6}$$

$$\frac{{\left( {16 \times 16 \times 16 \times 16{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$

On dividing individually 16 we always get a remainder 4.

$$\frac{{\left( {4 \times 4 \times 4 \times 4{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$

Required Remainder = 4

Note: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.