Number System: placement practice questions

Solution

The unit digits of powers of 9 repeat in a cycle of 2: 9, 1.

173 divided by 2 leaves remainder 1, so the unit digit is the same as that of 9¹, which is 9.

Solution

943 leaves remainder 7 and 347 leaves remainder 5 when divided by 9.

So the product leaves the same remainder as 7 × 5 = 35, which divided by 9 leaves 8.

Solution

13720 = 2³ × 5 × 7³.

Number of factors = product of (each power + 1) = 4 × 4 × 2 = 32.

Solution

Trailing zeros come from factors of 5 (there are always more 2s).

Count = ⌊95/5⌋ + ⌊95/25⌋ + ⌊95/125⌋ + … = 19 + 3 = 22.

Solution

A number is divisible by 11 when the difference between the sums of alternate digits is divisible by 11.

Trying x = 5 gives 247577, which is divisible by 11 (247577 ÷ 11 = 22507).

Solution

This is an arithmetic series with first term 4 and common difference 5.

Number of terms = (129 − 4) ÷ 5 + 1 = 26.

Sum = n × (first + last) ÷ 2 = 26 × 133 ÷ 2 = 1729.

Solution

The unit digits of powers of 17 repeat in a cycle of 4: 7, 9, 3, 1.

36 divided by 4 leaves remainder 0, so the unit digit is the same as that of 17⁴, which is 1.

Solution

872 leaves remainder 4 and 159 leaves remainder 5 when divided by 7.

So the product leaves the same remainder as 4 × 5 = 20, which divided by 7 leaves 6.

Solution

60 = 2² × 3 × 5.

Number of factors = product of (each power + 1) = 2 × 3 × 2 = 12.

Solution

Trailing zeros come from factors of 5 (there are always more 2s).

Count = ⌊229/5⌋ + ⌊229/25⌋ + ⌊229/125⌋ + … = 45 + 9 + 1 = 55.

Solution

A number is divisible by 11 when the difference between the sums of alternate digits is divisible by 11.

Trying x = 2 gives 105523, which is divisible by 11 (105523 ÷ 11 = 9593).

Solution

This is an arithmetic series with first term 1 and common difference 2.

Number of terms = (43 − 1) ÷ 2 + 1 = 22.

Sum = n × (first + last) ÷ 2 = 22 × 44 ÷ 2 = 484.

Solution

The unit digits of powers of 2 repeat in a cycle of 4: 2, 4, 8, 6.

154 divided by 4 leaves remainder 2, so the unit digit is the same as that of 2², which is 4.

Solution

542 leaves remainder 9 and 120 leaves remainder 3 when divided by 13.

So the product leaves the same remainder as 9 × 3 = 27, which divided by 13 leaves 1.

Solution

Trailing zeros come from factors of 5 (there are always more 2s).

Count = ⌊120/5⌋ + ⌊120/25⌋ + ⌊120/125⌋ + … = 24 + 4 = 28.

Solution

A number is divisible by 11 when the difference between the sums of alternate digits is divisible by 11.

Trying x = 5 gives 285780, which is divisible by 11 (285780 ÷ 11 = 25980).

Solution

This is an arithmetic series with first term 19 and common difference 7.

Number of terms = (236 − 19) ÷ 7 + 1 = 32.

Sum = n × (first + last) ÷ 2 = 32 × 255 ÷ 2 = 4080.

Solution

The unit digits of powers of 17 repeat in a cycle of 4: 7, 9, 3, 1.

135 divided by 4 leaves remainder 3, so the unit digit is the same as that of 17³, which is 3.

Solution

351 leaves remainder 1 and 680 leaves remainder 1 when divided by 7.

So the product leaves the same remainder as 1 × 1 = 1, which divided by 7 leaves 1.

Solution

504 = 2³ × 3² × 7.

Number of factors = product of (each power + 1) = 4 × 2 × 3 = 24.

Solution

Trailing zeros come from factors of 5 (there are always more 2s).

Count = ⌊42/5⌋ + ⌊42/25⌋ + ⌊42/125⌋ + … = 8 + 1 = 9.

Solution

A number is divisible by 11 when the difference between the sums of alternate digits is divisible by 11.

Trying x = 1 gives 732919, which is divisible by 11 (732919 ÷ 11 = 66629).

Solution

This is an arithmetic series with first term 8 and common difference 5.

Number of terms = (123 − 8) ÷ 5 + 1 = 24.

Sum = n × (first + last) ÷ 2 = 24 × 131 ÷ 2 = 1572.

Solution

The unit digits of powers of 13 repeat in a cycle of 4: 3, 9, 7, 1.

110 divided by 4 leaves remainder 2, so the unit digit is the same as that of 13², which is 9.

Solution

198 leaves remainder 3 and 769 leaves remainder 2 when divided by 13.

So the product leaves the same remainder as 3 × 2 = 6, which divided by 13 leaves 6.

Solution

7056 = 2⁴ × 3² × 7².

Number of factors = product of (each power + 1) = 5 × 3 × 3 = 45.

Solution

Trailing zeros come from factors of 5 (there are always more 2s).

Count = ⌊55/5⌋ + ⌊55/25⌋ + ⌊55/125⌋ + … = 11 + 2 = 13.

Solution

This is an arithmetic series with first term 3 and common difference 6.

Number of terms = (237 − 3) ÷ 6 + 1 = 40.

Sum = n × (first + last) ÷ 2 = 40 × 240 ÷ 2 = 4800.

Solution

The unit digits of powers of 3 repeat in a cycle of 4: 3, 9, 7, 1.

196 divided by 4 leaves remainder 0, so the unit digit is the same as that of 3⁴, which is 1.

Solution

669 leaves remainder 9 and 776 leaves remainder 6 when divided by 11.

So the product leaves the same remainder as 9 × 6 = 54, which divided by 11 leaves 10.