Numerical Aptitude

How to square a number ending with 5

This is a very simple method using which you could find the square of quite a few numbers which end with a five.

15²=225

25²=625

35²=1225

45²=2025

Method:

A number ending with 5 is of the form N=<A>5. To find the square of it i.e. N², just append 25 to the product of ( A ) * ( A + 1 ).

N=75 --> N=<A>5 --> so treat A as 7.
75² = the product (A) * (A + 1 ) appended with 25
75² = ( 7 * 8 )25
75² = 5625

N=95 --> N=<A>5 --> so treat A as 9.
95² = the product (A) * (A + 1 ) appended with 25
95² = ( 9 * 10 )25
95² = 9025

N=205 --> N=<A>5 --> so treat A as 20.
205² = the product (A) * (A + 1 ) appended with 25
205² = ( 20 * 21 )25
205² = 42025

How to square any two digit number (Closest multiple of 10 method)

Given the number N, first find its closest multiple of 10. Now find a number M which when added or subtracted with N becomes N's closest multiple of 10. To find the square of it i.e. N², just do ( N + M ) * ( N - M ) + M²

N=72, 72's closest multiple of 10 is 70 which is obtained when 2 is subtracted from 70. So M = 2.
N² = ( N + M ) * ( N - M ) + M²
72² = ( 72 + 2 ) * ( 72 - 2 ) + 2²
72² = ( 74 ) * ( 70 ) + 4
72² = 5180 + 4
72² = 5184

N=44, 44's closest multiple of 10 is 40 which is obtained when 4 is subtracted from 44. So M = 4.
N² = ( N + M ) * ( N - M ) + M²
44² = ( 44 + 4 ) * ( 44 - 4 ) + 4²
44² = ( 48 ) * ( 40 ) + 16
44² = 1920 + 16
44² = 1936

N=87, 87's closest multiple of 10 is 90 which is obtained when 3 is added to 87. So M = 3.
N² = ( N + M ) * ( N - M ) + M²
87² = ( 87 + 3 ) * ( 87 - 3 ) + 3²
87² = ( 90 ) * ( 84 ) + 9
87² = 7560 + 9
44² = 7569

Finding the units digit of an exponent.

What is the units digit of 7⁸⁵ ?

1

Wrong

3

Wrong

9

Wrong

7

Correct

Show / Hide Solution

The cyclicity of unit digit of 7ⁿ is a pattern of the form 7, 9, 3, 1 and the interval is 4. To find the unit digit of 7⁸⁷, do < exponent > mod < interval >. So we do 85 mod 4 which is 1. So the unit digit is 1st digit of the cyclic pattern i.e., 7

See below for detailed explanation.

How to find the units digit of a large exponent? Even with the use of calculators, its difficult to find the the digit in the unit's place of such large exponents. The catch here is to observe the unit's place of each of 7¹, 7², 7³, 7⁴, 7⁵, 7⁶, 7⁷...

71	72	73	74	75	76	77	78	...
7	49	343	2401	16807	117649	823543	5764801	...

We can observe that the unit digit follows a pattern and the pattern repeats itself at regular intervals of 4 digits i.e.,

7 9 3 1 7 9 3 1 ...

Therefore, in order to find the 'N'th power of 7, we have to do N mod 4.
If N mod 4 = 0, then the unit digit is 4th digit in the pattern ( 7 9 3 1 ) i.e., 1
If N mod 4 = 1, then the unit digit is 1st digit in the pattern ( 7 9 3 1 ) i.e., 7
If N mod 4 = 2, then the unit digit is 2nd digit in the pattern ( 7 9 3 1 ) i.e., 9
If N mod 4 = 3, then the unit digit is 3rd digit in the pattern ( 7 9 3 1 ) i.e., 3


Likewise the unit digit of a number for different powers is also found to be cyclic. The same has been calculated and summarised in the below table.
Observe that for base 1,5 and 6 the respective unit place raised to any power will always be 1, 5 and 6.
For 2, 3, 7, 8 the pattern repeats at intervals of 4.
For 4 and 9, the patter repeats at intervals of 2.

N1	N2	N3	N4	N5	N6	N7	N8	N9	Interval
1	1	1	1	1	1	1	1	1	1
2	4	8	6	2	4	8	6	2	4
3	9	7	1	3	9	7	1	3	4
4	6	4	6	4	6	4	6	4	2
5	5	5	5	5	5	5	5	5	1
6	6	6	6	6	6	6	6	6	1
7	9	3	1	7	9	3	1	7	4
8	4	2	6	8	4	2	6	8	4
9	1	9	1	9	1	9	1	9	2