Factors and Multiples: practice solutions, Grade 6 – download the PDF

1 Factors of 28 ★★★

Pairs: \(1 \times 28\), \(2 \times 14\), \(4 \times 7\). We stop because the next candidate, \(5\), does not divide \(28\), and \(6\) does not either.

The factors of \(28\) are \(1, 2, 4, 7, 14, 28\).

2 Multiples of 9 ★★★

\(9 \times 1 = 9\), \(9 \times 2 = 18\), \(9 \times 3 = 27\), \(9 \times 4 = 36\), \(9 \times 5 = 45\), \(9 \times 6 = 54\).

The first six multiples of \(9\) are \(9, 18, 27, 36, 45, 54\).

3 Prime or composite? ★★★

• \(17\): only the factors \(1\) and \(17\), so it is prime.
• \(21 = 3 \times 7\): composite.
• \(29\): not divisible by \(2, 3\) or \(5\) (and \(7 \times 7 = 49 > 29\)), so it is prime.
• \(39 = 3 \times 13\): composite.
• \(1\) has only one factor: neither.
• \(51 = 3 \times 17\): composite.

4 Divisible by 2, 5 or 10 ★★★

• \(340\) ends in \(0\): divisible by \(2\), \(5\) and \(10\).
• \(785\) ends in \(5\): divisible by \(5\) only.
• \(1{,}206\) ends in \(6\): divisible by \(2\) only.
• \(4{,}115\) ends in \(5\): divisible by \(5\) only.

5 True or false? ★★★

1. False. \(45 \div 7\) is not a whole number (\(7 \times 6 = 42\) and \(7 \times 7 = 49\)).
2. True. \(6\) is even, so \(6 \times n\) has \(2\) as a factor and is even.
3. False. \(1\) has only one factor, but a prime has exactly two.
4. False. \(2\) is prime and even.

6 Multiples in a range ★★★

\(7 \times 5 = 35\) is too small and \(7 \times 6 = 42\) works. Continue: \(7 \times 7 = 49\), \(7 \times 8 = 56\), \(7 \times 9 = 63\). Next is \(7 \times 10 = 70\), which is not less than \(70\).

The multiples are \(42, 49, 56, 63\).

7 GCF by listing ★★★

Factors of \(12\): \(1, 2, 3, 4, 6, 12\). Factors of \(30\): \(1, 2, 3, 5, 6, 10, 15, 30\).

Common factors: \(1, 2, 3, 6\). The greatest is \(6\), so \(\text{GCF}(12, 30) = 6\).

8 Digit sums ★★★

• \(2{,}718\): \(2 + 7 + 1 + 8 = 18\), divisible by \(3\) and by \(9\).
• \(5{,}436\): \(5 + 4 + 3 + 6 = 18\), divisible by \(3\) and by \(9\).
• \(7{,}305\): \(7 + 3 + 0 + 5 = 15\), divisible by \(3\) but not by \(9\).
• \(6{,}481\): \(6 + 4 + 8 + 1 = 19\), divisible by neither.

9 Factor tree of 84 ★★★

\(84 = 4 \times 21\), \(4 = 2 \times 2\) and \(21 = 3 \times 7\). All the end numbers are prime.

\(84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7\). Check: \(4 \times 3 \times 7 = 84\).

10 GCF with prime factors ★★★

\(56 = 2^3 \times 7\) and \(42 = 2 \times 3 \times 7\).

Common primes with the smaller exponent: \(2^1\) and \(7^1\). So \(\text{GCF}(56, 42) = 2 \times 7 = 14\).

11 LCM of 6 and 15 ★★★

Multiples of \(15\): \(15, 30, 45, \dots\) Multiples of \(6\): \(6, 12, 18, 24, 30, \dots\)

The first number on both lists is \(30\), so \(\text{LCM}(6, 15) = 30\).

12 Factoring out the GCF ★★★

1. \(\text{GCF}(45, 60) = 15\), so \(45 + 60 = 15(3 + 4) = 15 \times 7 = 105\).
2. \(\text{GCF}(24, 40) = 8\), so \(24 + 40 = 8(3 + 5) = 8 \times 8 = 64\).

13 Tiling a floor ★★★

The side must divide both \(30\) and \(18\), and we want the largest: \(\text{GCF}(30, 18) = 6\) (since \(30 = 2 \times 3 \times 5\) and \(18 = 2 \times 3^2\)).

Tiles per row: \(30 \div 6 = 5\). Rows: \(18 \div 6 = 3\). Total: \(5 \times 3 = 15\).

The largest tile has a side of \(6\) inches (about \(15\) cm) and \(15\) tiles are needed.

14 Two lighthouses ★★★

We want the smallest common multiple of \(9\) and \(12\). \(9 = 3^2\) and \(12 = 2^2 \times 3\), so \(\text{LCM} = 2^2 \times 3^2 = 36\).

They flash together again \(36\) seconds later. Check: \(36 = 9 \times 4 = 12 \times 3\).

15 A mystery number ★★★

A multiple of both \(4\) and \(7\) is a multiple of \(\text{LCM}(4, 7) = 28\). The multiples of \(28\) are \(28, 56, 84, \dots\)

Only \(56\) is between \(40\) and \(80\). Check: \(56 = 4 \times 14 = 7 \times 8\).

16 Find the mistake ★★★

\(96\) is a common multiple of \(8\) and \(12\), but it is not the least one. The multiples of \(12\) are \(12, 24, 36, \dots\) and \(24 = 8 \times 3\) is already a multiple of \(8\).

With primes: \(8 = 2^3\) and \(12 = 2^2 \times 3\), so \(\text{LCM} = 2^3 \times 3 = 24\). The product \(a \times b\) equals the LCM only when the numbers share no common factor other than \(1\).

17 School kits ★★★

\(72 = 2^3 \times 3^2\) and \(54 = 2 \times 3^3\), so \(\text{GCF}(72, 54) = 2 \times 3^2 = 18\).

She can make \(18\) kits, each with \(72 \div 18 = 4\) pencils and \(54 \div 18 = 3\) erasers.

18 Find the other number ★★★

Since \(\text{GCF} \times \text{LCM} = a \times b\), we get \(4 \times 60 = 12 \times b\), so \(b = 240 \div 12 = 20\).

Check: \(12 = 2^2 \times 3\) and \(20 = 2^2 \times 5\). \(\text{GCF} = 2^2 = 4\) and \(\text{LCM} = 2^2 \times 3 \times 5 = 60\). The other number is \(20\).

19 Three school bells ★★★

\(4 = 2^2\), \(6 = 2 \times 3\) and \(10 = 2 \times 5\). Take every prime with its largest exponent: \(\text{LCM} = 2^2 \times 3 \times 5 = 60\).

They ring together again after \(60\) minutes, at 1:00 p.m.

20 The missing digit ★★★

The digit sum is \(5 + \square + 4 = 9 + \square\), which must be a multiple of \(9\). With a single digit, \(\square = 0\) (sum \(9\)) or \(\square = 9\) (sum \(18\)). The numbers are \(504\) and \(594\).

Divisibility by \(4\): \(04\) is divisible by \(4\), but \(94 = 4 \times 23 + 2\) is not. So only \(504\) works (\(504 = 4 \times 126\)).

21 Sums of multiples ★★★

Write the numbers as \(6a\) and \(6b\) with \(a\) and \(b\) whole numbers. By the distributive property, \(6a + 6b = 6(a + b)\), a multiple of \(6\).

Example: \(42 = 6 \times 7\) and \(78 = 6 \times 13\), so \(42 + 78 = 6(7 + 13) = 6 \times 20 = 120\). As \(42 = 2 \times 3 \times 7\) and \(78 = 2 \times 3 \times 13\), \(\text{GCF}(42, 78) = 6\).

22 Hot dogs and buns ★★★

The number of items must be a multiple of \(8\) and of \(6\): \(\text{LCM}(8, 6) = 24\).

Hot dog packs: \(24 \div 8 = 3\). Bun packs: \(24 \div 6 = 4\). You buy \(3\) packs of hot dogs and \(4\) packs of buns, for \(24\) of each.