Q.3 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
Maximum number of columns in which the Army contingent and the band can march will be given by HCF (616, 32)
We can use Euclid's algorithm to find the HCF.
Step 1: since 616 > 32 so applying Euclid's division lemma to a= 616 and b= 32 we get integers q and r as 32 and 19
i.e 616 = 32 x 19 + 8
Step 2: since remainder r =8
0 so again applying Euclid's lemma to 32 and 8 we get integers 4 and 0 as the quotient and remainder
i.e 32 = 8 x 4 + 0
Step 3: Since remainder is zero so divisor at this stage will be the HCF
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
Concept Insight: In order to solve the word problems first step is to interpret the problem and identify what is to be determined. The key word "Maximum" means we need to find the HCF.Do not forget to write the unit in the answer.
Q.4 Use Euclid's division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Solution
Let a be any positive integer we need to prove that a2 is of the form 3m or 3m + 1 for some integer m.
Let b = 3 be the other integer so applying Euclid's division lemma to a and b=3
We get a = 3q + r for some integer q
0and r = 0, 1, 2
Therefore, a = 3q or 3q + 1 or 3q + 2
Now Consider a2
Where k1 = 3q2, k2 =3q2+2q and k3 = 3q2+4q+1 since q ,2,3,1 etc are all integers so is their sum and product.
So k1 k2 k3 are all integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1 for any integer m.
Concept Insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 3 because a must be of the form3m or 3m + 1. Do not forget to take a2. Note that variable is just a notation and not the absolute value.
Q.5 Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution
Let a be any positive integer and b = 3
a = 3q + r, where q
0 and 0
r < 3
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.
Concept Insight: In this problem Euclid's division lemma can be applied to integers a and b = 9 as well but using 9 will give us 9 values of r and hence as many cases so solution will be lengthy. Since every number which is divisible by 9 is also divisible by 3 so 3 is used.
Do not forget to take a3 and all the different values of a i.e
Solution
Maximum number of columns in which the Army contingent and the band can march will be given by HCF (616, 32)
We can use Euclid's algorithm to find the HCF.
Step 1: since 616 > 32 so applying Euclid's division lemma to a= 616 and b= 32 we get integers q and r as 32 and 19
i.e 616 = 32 x 19 + 8
Step 2: since remainder r =8
0 so again applying Euclid's lemma to 32 and 8 we get integers 4 and 0 as the quotient and remainder i.e 32 = 8 x 4 + 0
Step 3: Since remainder is zero so divisor at this stage will be the HCF
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
Concept Insight: In order to solve the word problems first step is to interpret the problem and identify what is to be determined. The key word "Maximum" means we need to find the HCF.Do not forget to write the unit in the answer.
Q.4 Use Euclid's division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
Solution
Let a be any positive integer we need to prove that a2 is of the form 3m or 3m + 1 for some integer m.
Let b = 3 be the other integer so applying Euclid's division lemma to a and b=3
We get a = 3q + r for some integer q
0and r = 0, 1, 2 Therefore, a = 3q or 3q + 1 or 3q + 2
Now Consider a2
Where k1 = 3q2, k2 =3q2+2q and k3 = 3q2+4q+1 since q ,2,3,1 etc are all integers so is their sum and product.
So k1 k2 k3 are all integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1 for any integer m.
Concept Insight: In order to solve such problems Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 3 because a must be of the form3m or 3m + 1. Do not forget to take a2. Note that variable is just a notation and not the absolute value.
Q.5 Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution
Let a be any positive integer and b = 3
a = 3q + r, where q
0 and 0 r < 3 Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1,
a3 = (3q +1)3
a3 = 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.
Concept Insight: In this problem Euclid's division lemma can be applied to integers a and b = 9 as well but using 9 will give us 9 values of r and hence as many cases so solution will be lengthy. Since every number which is divisible by 9 is also divisible by 3 so 3 is used.
Do not forget to take a3 and all the different values of a i.e
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