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HomeClass 10mathematicsChapter 1Exercise 1.1Q1
| Exercise 1.1

Question 1

Express each number as a product of its prime factors:

(i) 140140

(ii) 156156

(iii) 38253825

(iv) 50055005

(v) 74297429

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SOLUTION

UNDERSTAND THE QUESTION

The task is to write each given integer as a product of prime numbers. This requires repeatedly testing divisibility by the smallest possible prime (2, 3, 5, 7, …) and continuing the process with the quotient until only prime factors remain.

PART I

140
140

STEP 1 — CHECK DIVISIBILITY BY 2

140140 is even, so 140=2×70140 = 2 \times 70.

STEP 2 — FACTOR THE REMAINING EVEN NUMBER

7070 is also even, giving 70=2×3570 = 2 \times 35.

STEP 3 — FACTOR THE ODD PART

3535 ends with 5, so 35=5×735 = 5 \times 7, and both 5 and 7 are prime.

ANSWER

140=22×5×7
140 = 2^{2} \times 5 \times 7

PART II

156
156

STEP 1 — DIVIDE BY 2

156156 is even, so 156=2×78156 = 2 \times 78.

STEP 2 — DIVIDE THE QUOTIENT BY 2 AGAIN

7878 is even, giving 78=2×3978 = 2 \times 39.

STEP 3 — FACTOR THE ODD REMAINDER

3939 is divisible by 3 (sum of digits =12=12), so 39=3×1339 = 3 \times 13, and 13 is prime.

ANSWER

156=22×3×13
156 = 2^{2} \times 3 \times 13

PART III

3825
3825

STEP 1 — FACTOR OUT 5

The number ends in 5, so 3825=5×7653825 = 5 \times 765.

STEP 2 — FACTOR OUT ANOTHER 5

765765 also ends in 5, giving 765=5×153765 = 5 \times 153.

STEP 3 — FACTOR 153

153153 is divisible by 3 (sum of digits =9=9), so 153=3×51153 = 3 \times 51; 51=3×1751 = 3 \times 17, and 17 is prime.

ANSWER

3825=52×32×17
3825 = 5^{2} \times 3^{2} \times 17

PART IV

5005
5005

STEP 1 — TEST SMALL PRIMES

50055005 is odd and not divisible by 3 (digit sum =10=10). It ends with 5, so 5005=5×10015005 = 5 \times 1001.

STEP 2 — FACTOR 1001

10011001 is divisible by 7 (1001=7×1431001 = 7 \times 143).

STEP 3 — FACTOR 143

143=11×13143 = 11 \times 13, both of which are prime.

ANSWER

5005=5×7×11×13
5005 = 5 \times 7 \times 11 \times 13

PART V

7429
7429

STEP 1 — FIND A SMALL PRIME FACTOR

74297429 is not even, not divisible by 3 (digit sum =22=22) or 5. Testing 7, 11, 13 fails, but 7429÷17=4377429 \div 17 = 437, so 7429=17×4377429 = 17 \times 437.

STEP 2 — FACTOR 437

437÷19=23437 \div 19 = 23, giving 437=19×23437 = 19 \times 23, and both 19 and 23 are prime.

ANSWER

7429=17×19×23
7429 = 17 \times 19 \times 23

OVERALL FINAL ANSWER

(i) 
140=22×5×7
140 = 2^{2} \times 5 \times 7
(ii) 
156=22×3×13
156 = 2^{2} \times 3 \times 13
(iii) 
3825=52×32×17
3825 = 5^{2} \times 3^{2} \times 17
(iv) 
5005=5×7×11×13
5005 = 5 \times 7 \times 11 \times 13
(v) 
7429=17×19×23
7429 = 17 \times 19 \times 23

COMMON MISTAKES

Students often stop after finding one factor and forget to continue factoring the remaining quotient, leaving a composite number in the final product. Another frequent error is assuming a number is prime because it is not divisible by 2, 3, or 5; always test the next primes (7, 11, 13, …) until the quotient is prime.

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