__5 TIPS on cracking Aptitude Questions on Number Systems__

__Tip #1__: Factorize the expression

1. (a + b)(a - b) = (a^{2} - b^{2})

2. (a + b)^{2} = (a^{2} + b^{2} + 2ab)

3. (a - b)^{2} = (a^{2} + b^{2} - 2ab)

4. (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

5. (a^{3} + b^{3}) = (a + b)(a^{2} - ab + b^{2})

6. (a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2})

7. (a^{3} + b^{3} + c^{3} - 3abc) = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc – ac)

**Question: **297 x 297 =?

**Solution:**

297 x 297 = (300 – 3) (300 – 3) = (300 – 3)^{2} = 300^{2} + 3^{2} – 2x3x300

= **297 x 297 = 90000 + 9 – 1800 = 88209**

**Question: **__753 x 753 + 247 x 247 - 753 x 247__ = ?

753 x 753 x 753 + 247 x 247 x 247

**Solution: **

Let 753 = a and 247 = b. Then the fraction can be written as:

__a__^{2 }__+ b__^{2}__ – ab __= __ a__^{2}__ – ab + b__^{2}__ __ = __ 1 __ = __ 1 .__

a^{3} + b^{3} (a + b) (a^{2 }– ab + b^{2}) a + b **1000**

__Tip #2__: Use divisibility tests where applicable

**Question: **How many 3-digit numbers are completely divisible 6?

**Solution:**

*Smallest 3-digit number divisible by 6:*

100 is not divisible by 3.

101 is not divisible by 2.

102 is divisible by both 2 and 3 and hence, by 6.

*Largest 3-digit number divisible by 6:*

999 is not divisible by 2.

998 is not divisible by 3.

997 is not divisible by 2.

996 is divisible by both 2 and 3 and hence, divisible by 6.

Thus, the numbers form an AP with a = 102, d = 6, l = 996, n = ?

102 + (n – 1)6 = 996

= 6n = 996 – 102 + 6 = 900

= **n = 900 / 6 = 150 **

**Question: **If “481__d__673” is divisible by 9, what should be the smallest whole number in place of __d__?

**Solution:**

For the number to be divisible by 9, the sum of the digits should be divisible by 9.

4+8+1+6+7+3 = 29

The whole number just largest than 29 that is divisible by 9 is 36.

**Thus, **__d__ = 36 – 29 = 7

__Tip #3__: Apply the rules of Binomial Theorem

** **_{ n}** **

(x + y)^{n} = Σ ^{n}C_{r}x^{(n – r)}y^{r}

** **^{r=0}

1. (1 – x)^{-1} = 1 + x + x^{2} + x^{3} + … where x is between 0 and 1

2. (1 + x)^{ -1} = 1 – x + x^{2} – x^{3} + … where x is between 0 and 1

3. x^{n} + 1 is divisible by x + 1 for all odd values of n.

__Proof:__ Let us prove this formula using the method of induction.

When n=1, x^{n} + 1 = x+1 which is divisible by (x+1). When n=2, x^{2} + 1 is not divisible by (x+1).

Let x^{2k+1} + 1 be divisible by (x+1).

When n = 2k+3 (the next odd number after 2k + 1)

x^{2k+3} + 1 = x^{2}(x^{2k+1} + 1) + (1 - x^{2}) = x^{2}(x^{2k+1} + 1)+(1+x)(1-x), which is divisible by (x+1).

Thus, x^{n} + 1 is divisible by x + 1 when n is odd.

x^{n} – 1 is divisible by x + 1 only when n is even. [Proof is similar to the one above]

**Question: **What will be remainder when (67^{67} + 67) is divided by 68?

**Solution:**

From Binomial Theorem, we know, x^{n} + 1 is divisible by x + 1 only when n is odd.

67^{67} + 67 = (67^{67} + 1) + 66

**Thus, remainder = 66**

**Question: **Find the value of (0.999)^{3} correct to 3 decimal places.

**Solution:**

(0.999) ^{3} = (1 – 0.001)^{ 3}

= 1^{3} - C(3,1)1^{2}(0.001) + C(3,2)1(0.001)^{2} - (0.001)^{3}

= 1 - 0.003 = **0.997** (neglecting 3^{rd} and 4^{th} terms since they are « 0.001)

__Tip #4__: Remember the rules for division

Dividend = (Divisor x Quotient) + Remainder

Any recurring decimal can be written as the sum of a non-recurring decimal and a fraction of the recurring portion of the decimal out of (10^{n} – 1).

Ex: 0.125125125…= __0.125__ = 125/999 and 7.2341341341… = 7.2 + 341/9999

**Question: **Express 0.232323..... As a rational number.

**Solution:**

**0.232323… = 23/99.**

**Question: **A student mistook the divisor as 12 instead of 21 and obtained 35 as quotient and reminder as 0. What is the correct quotient?

**Solution: **

Let the dividend be x.

Then, x / 12 = 35

= x = 12 x 35

= ** Correct quotient = (12 x 35 / 21) = 4 x 5 = 20.**

**Question: **Find the smallest 6 digit number exactly divisible by 111.

**Solution: **

Smallest 6 digit number = 100000

When 100000 is divided by 111, the quotient is 900 and the remainder is 100.

The number would be divisible by 111 if the difference between 99900 (the number just lesser than 100000 that is divisible by 111) and the next divisible number is 111.

**Thus, the smallest 6 digit number divisible by 111 is 99900 + 111 = 100011.**

__Tip #5__: Memorize the formulae for sum of powers and apply where suitable

1. 1 + 2 + 3 + ….. + n = n (n + 1) / 2

2. 1^{2} + 2^{2} + 3^{2} + ….. + n^{2} = n (n + 1)(2n + 1) / 6

3. 1^{3} + 2^{3} + 3^{3} + …... + n^{3} = [ n (n + 1) / 2 ] ^{2}

**Question: **(51 + 52 + 53 + ... + 100) =?

**Solution: **

(51 + 52 + 53 + ….. + 100) = (1 + 2 + 3 + ….. + 100) – (1 + 2 + 3 + ….. + 50)

= __100(100 + 1)__ - __50(50 + 1)__

2 2

= 5050 – 1275

= **3775.**

**Question:** (2^{2} + 4^{2} + 6^{2} + ... + 20^{2}) =?

Solution:

(2^{2} + 4^{2} + 6^{2} + ... + 20^{2}) = 2^{2}(1^{2} + 2^{2} + 3^{2} + ….. + 10^{2})

= 2^{2} x __10(10 + 1) (2•10 + 1)__

6

= 4 x 10 x 11 x 21 / 6

=** 1540.**