Express each number as a product of its prime factors:
(i) 140,
(ii) 156,
(iii) 3825,
(iv) 5005,
(v) 7429
(i) \(2^2 \times 5 \times 7\)
(ii) \(2^2 \times 3 \times 13\)
(iii) \(3^2 \times 5^2 \times 17\)
(iv) \(5 \times 7 \times 11 \times 13\)
(v) \(17 \times 19 \times 23\)
We need to express each number as a product of prime numbers only. We will divide the given number, step by step, by the smallest prime factor each time and continue until all the remaining numbers are prime.
(i) 140
First, look at 140. It is even, so it is divisible by \(2\), which is a prime number.
Divide 140 by \(2\): \(140 \div 2 = 70\).
Now we have 70. This is also even, so again it is divisible by \(2\).
Divide 70 by \(2\): \(70 \div 2 = 35\).
Now we have 35. It is not even, so it is not divisible by \(2\). Next check if it is divisible by the prime \(3\). A number is divisible by 3 if the sum of its digits is divisble by 3. The sum of digits of 35 is \(3 + 5 = 8\), which is not divisible by \(3\), so 35 is not divisible by \(3\).
Try the next prime, \(5\). The last digit of 35 is 5, so it is divisible by \(5\).
Divide 35 by \(5\): \(35 \div 5 = 7\).
Now we get 7. The number 7 is a prime number.
So the prime factors of 140 are \(2, 2, 5, 7\). Writing them as a product, we get \(2 \times 2 \times 5 \times 7\).
In index form, \(2 \times 2 = 2^2\), so we write: \(140 = 2^2 \times 5 \times 7\).
(ii) 156
Look at 156. It is even, so it is divisible by \(2\).
Divide 156 by \(2\): \(156 \div 2 = 78\).
Now we have 78. This is also even, so again it is divisible by \(2\).
Divide 78 by \(2\): \(78 \div 2 = 39\).
Now we have 39. It is not even, so not divisible by \(2\). Check divisibility by \(3\). The sum of digits of 39 is \(3 + 9 = 12\), and 12 is divisible by \(3\), so 39 is divisible by \(3\).
Divide 39 by \(3\): \(39 \div 3 = 13\).
Now we get 13. The number 13 is a prime number.
So the prime factors of 156 are \(2, 2, 3, 13\). As a product, \(156\) is \(2 \times 2 \times 3 \times 13\).
We can write \(2 \times 2 = 2^2\), so we get \(2^2 \times 3 \times 13\).
(iii) 3825
Start with the smallest prime, \(2\). Since 3825 is odd, it is not divisible by \(2\). So we will not check \(2\) again for any of the later numbers.
Next prime is \(3\). The sum of digits is \(3 + 8 + 2 + 5 = 18\), and 18 is divisible by \(3\). So 3825 is divisible by \(3\).
Divide 3825 by \(3\): \(3825 \div 3 = 1275\).
Now we continue checking with the same prime, \(3\). The sum of digits of 1275 is \(1 + 2 + 7 + 5 = 15\), which is divisible by \(3\). So 1275 is divisible by \(3\).
Divide 1275 by \(3\): \(1275 \div 3 = 425\).
Now check 425 for \(3\). The sum of digits is \(4 + 2 + 5 = 11\), which is not divisible by \(3\). So we will not check \(3\) again for any later numbers.
The next prime is \(5\). The last digit of 425 is 5, so 425 is divisible by \(5\).
Divide 425 by \(5\): \(425 \div 5 = 85\).
Continue checking with \(5\). The last digit of 85 is 5, so 85 is divisible by \(5\).
Divide 85 by \(5\): \(85 \div 5 = 17\).
Now check 17 for \(5\). The last digit is 7, so it is not divisible by \(5\). So we will not check \(5\) again for further numbers.
The next prime is \(7\). 17 is not divisible by \(7\), and since \(7 > \sqrt{17}\), 17 must be prime.
Thus, the prime factors of 3825 are \(3, 3, 5, 5, 17\).
As a product, we write this as \(3 \times 3 \times 5 \times 5 \times 17\).
Grouping equal primes: \(3^2 \times 5^2 \times 17\).
(iv) 5005
Take 5005. The last digit is 5, so it is divisible by \(5\).
Divide 5005 by \(5\): \(5005 \div 5 = 1001\).
Now consider 1001. It is not even, and the sum of its digits is \(1 + 0 + 0 + 1 = 2\), so it is not divisible by \(2\) or \(3\). Check for \(5\) next: it does not end in 0 or 5, so it is not divisible by \(5\).
Try the next prime, \(7\). We test and find that \(1001 \div 7 = 143\), so 1001 is divisible by 7.
Now we have 143. It is not even, and the sum of digits is \(1 + 4 + 3 = 8\), so it is not divisible by \(2\) or \(3\). It does not end in 0 or 5, so not divisible by \(5\).
Try the next prime, \(7\), but \(143 \div 7\) is not a whole number.
Try \(11\). We find that \(143 \div 11 = 13\).
Now we get 13, which is prime.
So the prime factors of 5005 are \(5, 7, 11, 13\).
We write this as a product: \(5 \times 7 \times 11 \times 13\).
(v) 7429
Now take 7429. It is not even, so it is not divisible by \(2\). Check the sum of digits: \(7 + 4 + 2 + 9 = 22\). Since 22 is not divisible by \(3\), 7429 is not divisible by \(3\).
The last digit is 9, so it is not divisible by \(5\).
Try dividing by the prime \(7\). The value \(7429 \div 7\) is not a whole number, so 7 is not a factor.
Try the next prime, \(11\). The division \(7429 \div 11\) is also not a whole number.
Try \(13\). Again, \(7429 \div 13\) does not give a whole number.
Next, try the prime \(17\). We get \(7429 \div 17 = 437\), which is a whole number, so 17 is a factor.
Now consider 437. It is not even, and the sum of its digits is \(4 + 3 + 7 = 14\), so it is not divisible by \(3\). It does not end in 0 or 5, so not divisible by \(5\).
Try dividing 437 by \(17\); it does not give a whole number.
Try the next prime, \(19\). We get \(437 \div 19 = 23\), which is a whole number.
Now we have 23, which is a prime number.
So the prime factors of 7429 are \(17, 19, 23\).
Writing them as a product, we get \(17 \times 19 \times 23\).