Use app

Properties of Multiples

Q: Check whether \$6^n\$ can end with the digit \$0\$ for any natural number \$n\$.

If any digit has the last digit \$10\$ that means it divisible by \$10.\$The factor of \$10 = 2 \times 5,\$ So value of \$6^n\$ should be divisible by \$2\$ and \$5.\$Both \$6^n\$ is divisible by \$2\$ but not divisible by \$5.\$So, it can not end with \$0.\$

Q: Find the greatest number that will divide \$43, 91\$ and \$183\$ so as to leave the same remainder in each case.

Here, we need to find differences between the given numbers. If two numbers give the same remainder when divided by some other number, then their difference must give a remainder of zero when divided by that number.Our numbers here are \$91-43=48, 183-91=92, 183-43=140\$ So we have the set of numbers \$\{48, 92, 140\}\$ and we want to know the biggest number that divides all these numbers. So,\$48=2 \times 2 \times 2\times 3\$\$92=2 \times 2 \times 23\$\$140=2 \times 2 \times 5\times 7\$The greatest common divisor of \$\{48, 92, 140\}\$ is \$4.\$ So, answer is \$4.\$

Q: Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) \$12, 15\$ and \$21\$ (ii) \$17, 23\$ and \$29\$ (iii) \$8, 9\$ and \$25\$

Using prime factorisation method:(i) 12, 15 and 21Factor of \$12 = 2 \times 2 \times 3\$Factor of \$15 = 3 \times 5\$Factor of \$21 = 3 \times 7\$HCF \$(12, 15, 21) = 3\$LCM \$(12, 15, 21) = 2 \times 2 \times 3 \times 5 \times 7 = 420\$(ii) 17, 23 and 29Factor of \$17 = 1 \times 17\$Factor of \$23 = 1 \times 23\$Factor of \$29 = 1 \times 29\$HCF \$(17, 23, 29) = 1\$LCM \$(17, 23, 29) = 1 \times 17 \times 23 \times 29 = 11,339\$(iii) 8, 9 and 25Factor of \$8 = 2 \times 2 \times 2 \times 1\$Factor of \$9 = 3 \times 3 \times 1\$Factor of \$25 = 5 \times 5 \times 1\$HCF \$(8, 9, 25) = 1\$LCM \$(8, 9, 25) = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 1,800\$

Q: Express each number as a product of its prime factors:(i) \$140\$ (ii) \$156\$ (iii) \$3825\$ (iv) \$5005 \$ (v) \$7429\$

(i) \$140\$\$140 = 2 \times 70\$\$= 2 \times 2 \times 35\$\$= 2 \times 2 \times 5 \times 7\$\$= 2 \times 2 \times 5 \times 7 \times 1\$(ii) \$156\$\$156 = 2 \times 78\$\$= 2 \times 2 \times 39\$\$= 2 \times 2 \times 3 \times 13\$\$= 2 \times 2 \times 3 \times 13 \times 1\$(iii) \$3825\$\$3825 = 3 \times 1275\$\$= 3 \times 3 \times 425\$\$= 3 \times 3 \times 5 \times 85\$\$= 3 \times 3 \times 5 \times 5 \times 17\$\$= 3 \times 3 \times 5 \times 5 \times 17 \times 1\$(iv) \$5005\$\$5005 = 5 \times 1001\$\$= 5 \times 7 \times 143\$\$= 5 \times 7 \times 11 \times 13\$\$= 5 \times 7 \times 11 \times 13 \times 1\$(v) \$7429\$\$7429 = 17 \times 437\$\$= 17 \times 19 \times 23\$\$= 17 \times 19 \times 23 \times 1\$

Q: Find the first four common multiples of the following : \$3\$ and \$4\$.

Multiples of \$ 3 = 3, 6, 9, 12, 15, 18.. \$Multiples of \$ 4 = 4, 8, 12, 16, 20.. \$The first common multiple will be \$ 12 \$And the next common multiples will be multiples of \$ 12 \$Hence, first four common multiples of \$ 3, 4 \$ are \$ 12, 24, 36, 48 \$

Q: Find the number of odd and even divisors of \$600\$

\$600=2\times 3\times 2\times 2\times 5\times 5={2}^{3}\times{3}^{1}\times{5}^{2}\$Number of divisors\$=\left(3+1\right)\left(1+1\right)\left(2+1\right)=4\times 2\times 3=24\$Number of odd divisors\$=\left(1+1\right)\left(2+1\right)=2\times 3=6\$Number of even divisors\$=24-6=18\$

Q: Fill in the blanks :Every number is a multiple of .............

Every number is a multiple of itself

Q: A number \$n\$ is said to be 'perfect' if the sum of all its divisors (excluding \$n \$ itself) is equal to \$n\$. An example of a perfect number is

Lets check each option one by one A. \$9\$ divisor are \$1\$ and \$3\$. \$1+3=4\neq 9\$ so not a perfect number.B. \$15\$ divisors are \$1, 3,\$ and \$5\$. \$1+3+5=9\neq 15\$ so not a perfect numberC. \$21\$ divisors are 1, 3 and 7. \$1+3+7 = 11\neq 21\$, so not a perfect number.D. \$6\$ divisors are \$1,2\$ and \$3\$. \$1+2+3 =6\$ It is a perfect number So correct answer is option D

Q: A group of \$616\$ students is to march behind an army band of \$32\$ members in a parade. The two groups must march in the same number of columns. What can be the maximum number of columns in which they march?

Maximum number of columns which is common to both the groups = HCF (616, 32)HCF (616, 32):616 = (32 x 19) + 832 = (8 x 4) + 0Remainder in this case is zero.∴ HCF (616, 32) = 8

Maths

definition

Lowest Common Multiple

LCM of two numbers is the smallest whole number which is a multiple of both the numbers.
The lowest common multiple is found by multiplying all the factors which appear in either (prime) factor list of both numbers.
Multiples of

12

are

12, 24, 36, 48, 60, 72, 84, 96 . . .

12 = 1 \times 2 \times 2 \times 3

Multiples of

16

are

16, 32, 48 . . .

16 = 1 \times 2 \times 2 \times 2 \times 2

The LCM of

12

and

16

48

1 \times 2^{4} \times 3 = 48

REVISE WITH CONCEPTS

Properties of Factors

ExampleDefinitionsFormulaes

Quick Summary With Stories