Solution : We have,  $$7 \times 11 \times 13$$ + 13 = 1001 + 13 =1014 1014 = $$2 \times 3 \times 13 \times 13$$ So, it is the product of more than two prime numbers. 2, 3 and 13. Hence, it is a composite number. $$7 \times 6 \times 5 \times 4 \times 3 … ## Check whether \(6^n$$ can end with the digit 0 or any n $$\in$$ N.

Solution : If the number $$6^n$$ ends with the digit zero. Then it is divisible by 5. Therefore the prime factorisation of $$6^n$$ contains the prime 5. This is not possible because the only primes in the factorisation of $$6^n$$ are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that …

## Given H.C.F. (306, 657) = 9, find L.C.M. (306, 657).

Solution : We have, H.C.F. (306, 657) = 9. We know that, Product of L.C.M and H.C.F = Product of two numbers. $$\implies$$  L.C.M $$\times$$ 9 = $$306 \times 657$$ $$\implies$$  L.C.M = $$306 \times 657 \over 9$$ = 22338 Hence, L.C.M is 22338.

## Find the H.C.F and L.C.M of the following integers by applying prime factorisation method.

Question : Find the H.C.F and L.C.M of the following integers by applying prime factorisation method. (i)  12, 15, 21 (ii)  17, 23, 29 (iii)  8, 9 and 25 Solution : (i) 12, 15, 21 12 = $$2 \times 2 \times 3$$ 15 = $$3 \times 5$$ 21 = $$3 \times 7$$ Here 3 is a …

## Find the L.C.M and H.C.F of the following pairs of integers and verify :

Question : Find the L.C.M and H.C.F of the following pairs of integers and verify : L.C.M $$\times$$ H.C.F = Product of the two numbers (i)  26 and 91 (ii)  510 and 92 (iii)  336 and 54 Solution :  (i)  26 and 91 26 = 2 $$\times$$ 13        and       91 …

## Express each number as a product of its prime factors :

Question : Express each number as a product of its prime factors : (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution : (i)  We use the division method as shown below : $$\therefore$$   140 = $$2 \times 2 \times 5 \times 7$$ = $$2^2 \times 5 \times 7$$ (ii)  We use the division …

## Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m or 9m + 1 or 9m + 8.

Solution : Let m be any positive integer. Then it is of the form 3m, 3m + 1 or 3m + 2. Now, we have to prove that the cube of these can be rewritten in the form 9q, 9q + 1 or 9q + 8. Now, $$(3m)^3$$ = $$27m^3$$ = $$9(m^3)$$ = 9q, where …

## Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution : By Euclid’s Division Algorithm, we have a = bq + r       …………..(i) On putting b = 3 in (1), we get a = 3q + r,      [0 $$\le$$ r < 3] If r = 0   a = 3q  $$\implies$$  $$a^2$$ = $$9q^2$$                …

## An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?

Solution : To find the maximum number of columns, we have to find the H.C.F. of 616 and 32 i.e. 616 = 32 $$\times$$ 19 + 8 and 32 = 8 $$\times$$ 4 + 0 $$\therefore$$ H.C.F of 616 and 32 is 8. Hence, maximum number of columns is 8.