Question 27

We will break down each number into a multiplication of only prime numbers.

Step 1 — Prime factors of 64

Let us start with the number 64. We divide 64 by the smallest prime number, which is 2. We keep dividing the result by 2 until we cannot anymore. We stop when the result is 1.

$64 \div 2 = 32$

$32 \div 2 = 16$

$16 \div 2 = 8$

$8 \div 2 = 4$

$4 \div 2 = 2$

$2 \div 2 = 1$

So, 64 is a multiplication of these prime numbers.

$\boxed{64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

Step 2 — Prime factors of 104

Now let us find factors for 104. We divide 104 by the smallest prime number, 2. We keep dividing by 2 until we cannot anymore. Then we try the next prime number.

$104 \div 2 = 52$

$52 \div 2 = 26$

$26 \div 2 = 13$

The number 13 is a prime number. We divide 13 by 13.

$13 \div 13 = 1$

So, 104 is a multiplication of these prime numbers.

$\boxed{104 = 2 \times 2 \times 2 \times 13}$

Step 3 — Prime factors of 105

Let us find factors for 105. 105 cannot be divided by 2. We check if 105 can be divided by 3. The sum of digits $1+0+5 = 6$. Since 6 can be divided by 3, 105 can be divided by 3.

$105 \div 3 = 35$

35 cannot be divided by 3. We check if 35 can be divided by 5. Since 35 ends in 5, it can be divided by 5.

$35 \div 5 = 7$

The number 7 is a prime number. We divide 7 by 7.

$7 \div 7 = 1$

So, 105 is a multiplication of these prime numbers.

$\boxed{105 = 3 \times 5 \times 7}$

Step 4 — Prime factors of 243

Let us find factors for 243. 243 cannot be divided by 2. We check if 243 can be divided by 3. The sum of digits $2+4+3 = 9$. Since 9 can be divided by 3, 243 can be divided by 3.

$243 \div 3 = 81$

$81 \div 3 = 27$

$27 \div 3 = 9$

$9 \div 3 = 3$

$3 \div 3 = 1$

So, 243 is a multiplication of these prime numbers.

$\boxed{243 = 3 \times 3 \times 3 \times 3 \times 3}$

Step 5 — Prime factors of 320

Let us find factors for 320. We divide 320 by the smallest prime number, 2. We keep dividing by 2 until we cannot anymore.

$320 \div 2 = 160$

$160 \div 2 = 80$

$80 \div 2 = 40$

$40 \div 2 = 20$

$20 \div 2 = 10$

$10 \div 2 = 5$

The number 5 is a prime number. We divide 5 by 5.

$5 \div 5 = 1$

So, 320 is a multiplication of these prime numbers.

$\boxed{320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5}$

Step 6 — Prime factors of 141

Let us find factors for 141. 141 cannot be divided by 2. We check if 141 can be divided by 3. The sum of digits $1+4+1 = 6$. Since 6 can be divided by 3, 141 can be divided by 3.

$141 \div 3 = 47$

The number 47 is a prime number. We divide 47 by 47.

$47 \div 47 = 1$

So, 141 is a multiplication of these prime numbers.

$\boxed{141 = 3 \times 47}$

Step 7 — Prime factors of 1728

Let us find factors for 1728. We divide 1728 by the smallest prime number, 2. We keep dividing by 2 until we cannot anymore.

$1728 \div 2 = 864$

$864 \div 2 = 432$

$432 \div 2 = 216$

$216 \div 2 = 108$

$108 \div 2 = 54$

$54 \div 2 = 27$

27 cannot be divided by 2. We check if 27 can be divided by 3. The sum of digits $2+7 = 9$. Since 9 can be divided by 3, 27 can be divided by 3.

$27 \div 3 = 9$

$9 \div 3 = 3$

$3 \div 3 = 1$

So, 1728 is a multiplication of these prime numbers.

$\boxed{1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3}$

Step 8 — Prime factors of 729

Let us find factors for 729. 729 cannot be divided by 2. We check if 729 can be divided by 3. The sum of digits $7+2+9 = 18$. Since 18 can be divided by 3, 729 can be divided by 3.

$729 \div 3 = 243$

$243 \div 3 = 81$

$81 \div 3 = 27$

$27 \div 3 = 9$

$9 \div 3 = 3$

$3 \div 3 = 1$

So, 729 is a multiplication of these prime numbers.

$\boxed{729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3}$

Step 9 — Prime factors of 1024

Let us find factors for 1024. We divide 1024 by the smallest prime number, 2. We keep dividing by 2 until we cannot anymore.

$1024 \div 2 = 512$

$512 \div 2 = 256$

$256 \div 2 = 128$

$128 \div 2 = 64$

$64 \div 2 = 32$

$32 \div 2 = 16$

$16 \div 2 = 8$

$8 \div 2 = 4$

$4 \div 2 = 2$

$2 \div 2 = 1$

So, 1024 is a multiplication of these prime numbers.

$\boxed{1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

Step 10 — Prime factors of 1331

Let us find factors for 1331. 1331 cannot be divided by 2, 3, 5, or 7. We check if 1331 can be divided by 11.

$1331 \div 11 = 121$

$121 \div 11 = 11$

$11 \div 11 = 1$

So, 1331 is a multiplication of these prime numbers.

$\boxed{1331 = 11 \times 11 \times 11}$

Step 11 — Prime factors of 1000

Let us find factors for 1000. We divide 1000 by the smallest prime number, 2. We keep dividing by 2 until we cannot anymore.

$1000 \div 2 = 500$

$500 \div 2 = 250$

$250 \div 2 = 125$

125 cannot be divided by 2 or 3. We check if 125 can be divided by 5. Since 125 ends in 5, it can be divided by 5.

$125 \div 5 = 25$

$25 \div 5 = 5$

$5 \div 5 = 1$

So, 1000 is a multiplication of these prime numbers.

$\boxed{1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5}$

Answer

(i) $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$ (ii) $104 = 2 \times 2 \times 2 \times 13$ (iii) $105 = 3 \times 5 \times 7$ (iv) $243 = 3 \times 3 \times 3 \times 3 \times 3$ (v) $320 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$ (vi) $141 = 3 \times 47$ (vii) $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$ (viii) $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$ (ix) $1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ (x) $1331 = 11 \times 11 \times 11$ (xi) $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$