Number theory book

Number theory book is a significant part of this book, even though it constitutes a small portion of the exam, because many of the most challenging problems involve this subject.

This page features advanced Number Theory problems commonly seen on the most challenging ACT math book questions. The material is designed for students aiming for top scores and is adapted from my book.These problems focus on patterns, divisibility, prime numbers, and number properties that frequently appear in high-difficulty ACT exams.You can explore more detailed explanations in the full book here ACT Math book on Amazon.

Free ACT number theory book Practice Questions:

Digit in Decimal Representation

Consider the problem of finding the $100^{th}$ digit after the decimal point in the decimal representation of $\dfrac{1}{7}$. Entering $\dfrac{1}{7}$ into a calculator yields $0.\overline{142857}$.

Notice that the pattern repeats every 6 digits. Exam problems often involve division by 7, resulting in a cycle of 6 digits, though other fractions may have different cycle lengths. Usually, the repeating pattern is given, but it may need to be discovered by dividing.

Returning to the example, to find the digit, divide the target position by the cycle length. For the $100^{th}$ digit with a cycle of 6, we calculate the remainder of $100 \div 6$. This can be done by hand or by multiplying the decimal part of the calculator result ($0.666\dots$) by the divisor (6) to obtain the remainder 4. Therefore, the answer is the fourth digit to the right of the decimal in the repeating pattern, which is 8.

Note that when computing the remainder in the division step, some calculators might be able to provide the remainder directly.

1. What is the $17^{th}$ digit to the right of the decimal point in the decimal expansion of $\dfrac{7}{11}$?3

$ 1 $
$ 2 $
$ 3 $
$ 6$

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2. What is the $88^{th}$ digit to the right of the decimal point in the decimal representation of $\dfrac{241}{999}$? 4

$ 1 $
$ 2 $
$3$
$ 4$

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3. What is the $125^{th}$ digit after the decimal in the decimal representation of $\dfrac{6}{7}$? 4

$ 2 $
$4$
$ 6$
$ 7 $

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4. What is the product of the $100^{th}$ and $200^{th}$ digits in the decimal representation of $\dfrac{4}{7}$? 5

$8$
$ 16 $
$ 28 $
$35$

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Units Digit of a Number to a Power

Finding the units digit of a number raised to a large power involves identifying the cyclic pattern of the powers. For example, calculating the units digit of $132^{87}$ is equivalent to finding the units digit of $2^{87}$, because digits in the tens place and higher do not affect the units digit of the result.

We examine the units digits of the powers of 2: $$ 2^1=2, \quad 2^2=4, \quad 2^3=8, \quad 2^4=6, \quad 2^5=2. $$ Since $2^5$ has the same units digit as $2^1$, the pattern repeats every 4 powers. This periodicity of 4 is common in units digit problems, similar to the powers of $i$ (where $i=\sqrt{-1}$).

To solve the problem, we determine the remainder when the exponent 87 is divided by the cycle length 4. Dividing 87 by 4 yields a remainder of 3. Alternatively, using a calculator: $$ \frac{87}{4} = 21.75 \implies 0.75 \cdot 4 = 3. $$ Since the remainder is 3, the units digit of $132^{87}$ is the same as that of $2^3$. Because $2^3=8$, the units digit is 8.

5. The digit in the ones place of $8^{127}$ is $2$. What is the digit in the ones place of $8^{130}$? 4

$2$
$4$
$6$
$ 8 $

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6. What is the units digit of $38^{39}$? 5

$ 2 $
$4 $
$ 6 $
$8$

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7. What is the units digit of $53^{250}$? 5

$ 1 $
$3$
$ 7 $
$ 9$

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Solutions

1. (D) $\dfrac{7}{11} = 0.\overline{63}$. The digits in odd positions are 6.

2. (B) $\dfrac{241}{999}=0.\overline{241}$. The decimal repeats every 3 digits. $88\mod3=1$, so we take the first digit, which is 2.

3. (B) $\dfrac{6}{7}=0.\overline{857142}$. The decimal repeats every 6 digits. You can determine that with a calculator. $125\mod6=5$. The $5^{th}$ digit is a 4, so the answer is 4.

4. (C) $\dfrac{4}{7}$ is $0.\overline{571428}$. The pattern repeats every $6$ digits. $100\mod6=4$ and $200\mod6=2$. The fourth digit is 4 and the second digit is 7. We take the product of those digits, $4\cdot7=28$.

5. (B) Multiply the given units digit (2) by the difference in powers: $2\cdot 8^{130 - 127}=2\cdot 8^3=2\cdot 512=1024 \equiv 4 \pmod{10}$.

6. (A) $8^1 \equiv 8 \pmod{10}$, $8^2 \equiv 4 \pmod{10}$, $8^3 \equiv 2 \pmod{10}$, $8^4 \equiv 6 \pmod{10}$, $8^5=8\mod10$, so the pattern repeats every 4. Since $39 \equiv 3 \pmod 4$, $38^{39} \equiv 38^3 \equiv 8^3 \equiv 2 \pmod{10}$.

7. (D) $3^1 \equiv 3 \pmod{10}$, $3^2 \equiv 9 \pmod{10}$, $3^3 \equiv 7 \pmod{10}$, $3^4 \equiv 1 \pmod{10}$, so the pattern repeats every 4. Since $250 \equiv 2 \pmod 4$, the units digit corresponds to $3^2=9$.

8. (D) Positive and odd. An odd integer times an even integer is an even integer. A negative number raised to an even integer power is positive. An odd number raised to an integer power is odd.

9. (B) $130 \div 24 = 5$ remainder $10$. Add the 10 remaining hours to 5 PM: $5 + 10 = 15$, which corresponds to 3 AM.

10. (D) For $6\cdot 6$, Perimeter $=2\cdot 6+2\cdot 6=24$; for $9\cdot 4$, Perimeter $=2\cdot 9+2\cdot 4=26$; for $12\cdot 3$, Perimeter $=2\cdot 12+2\cdot 3=30$; for $2\cdot 18$, Perimeter $=2\cdot 2+2\cdot 18=40$; for $1 \cdot 36$, Perimeter $=2\cdot 1+2\cdot 36=74$.

11. (A) For 9, the factors are 1, 3, and 9, which is odd. For 10, the factors are 1, 2, 5, and 10, which is even. For 12, the factors are 1, 2, 3, 4, 6, and 12, which is even. For 18, the factors are 1, 2, 3, 6, 9, and 18, which is even. In general, only perfect squares have an odd number of distinct factors.

12. (C) $2x+3x=5x$. $5x$ must be divisible by $5$ if $x$ is an integer.

13. (D) We only have to look at the last digit because if, for example, $7^2=49$, the perfect square of any number ending in 7 will end in 9. $$ \text{$1^2=1$, $2^2=4$, $3^2=9$, $4^2=16$, $5^2=25$, $6^2=36$, $7^2=49$, $8^2=64$, $9^2=81$, $0^2=0,$} $$ so a perfect square must end in 0, 1, 4, 5, 6, or 9. Therefore, a square cannot end in 2, 3, 7, or 8.

14. (C) Taking numbers whose units digit is double their tens digit, we get 12, 24, 36, and 48. There can be no more because twice 5 or greater is not a single digit. Therefore, the answer is 4.

15. (B) Take all the numbers in that range not divisible by 2, 3, or 5. A number is divisible by 3 if and only if its digits add up to a multiple of 3. A number is divisible by 5 if and only if it ends in 5 or 0. Therefore, we have reduced the list of possible primes to 101, 103, 107, 109, 113, 119, 121, and 127. Note that 119 is divisible by 7 and 121 is divisible by 11. We do not need to check for divisibility by primes 13 or higher, since $13^2=169>130$. Therefore, the only primes in the range are 101, 103, 107, 109, 113, and 127, so there are 6 primes.

16. (B) Take 1 minus the proportion of numbers without a 3 to determine the proportion with a 3. $9/10$ of units digits do not have a 3. However, $8/9$ of tens digits lack a 3 because there are only 9 possible tens digits ($1\text{--}9$). $$ 1-\dfrac{8}{9}\cdot\dfrac{9}{10}=1-\dfrac{8}{10}=\dfrac{1}{5}. $$ Another approach is to enumerate the numbers with a 3: 13, 23, 30--39, 43, 53, 63, 73, 83, 93. There are 18 numbers with 3s, so $\dfrac{18}{90}=\dfrac{1}{5}$. A third approach is to count the numbers without a 3. There are 8 choices for the tens digit and 9 choices for the units digit, so there are $8 \cdot 9 = 72$ numbers without a 3. Subtracting from the total gives $90-72=18$ numbers with a 3, so the fraction is $\dfrac{18}{90}=\dfrac{1}{5}$.

17. (C) In this case, the most practical approach is to take 1 minus the proportion of numbers without a 7. $9/10$ of the units and tens digits do not have a 7. However, $8/9$ of the hundreds digits lack a 7 because there are only 9 possible hundreds digits ($1\text{--}9$). $$ 1-\dfrac{8}{9}\cdot\dfrac{9}{10}\cdot\dfrac{9}{10}=1-\dfrac{72}{100}=\dfrac{28}{100}=\dfrac{7}{25}. $$ Another approach is to count the numbers without a 7. There are 8 choices for the hundreds digit and 9 choices for the tens and units digits, so there are $8 \cdot 9 \cdot 9 = 648$ numbers without a 7. Subtracting from the total ($900$) gives $252$ numbers with a 7, so the fraction is $\dfrac{252}{900}=\dfrac{7}{25}$.

18. (D) Solution 1: The last two digits of the first 25 perfect squares are: $$ 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25. $$ After $25^2$, the digits repeat symmetrically (e.g., $26^2$ ends like $24^2$). After $50^2$, the digits repeat normally (e.g., $51^2$ ends like $1^2$). So, looking at the list above, 84 is the only answer choice listed. Solution 2: Eliminate choices ending in 2, 3, 7, and 8, which cannot be perfect squares (as shown in Problem 13). This eliminates 18 and 32 (choices A and B). The remaining choices (54 and 84) are even numbers. Every perfect square which is divisible by 2 must also be divisible by 4. Using the criteria for divisibility by 4 (a number is divisible by 4 if and only if the ending 2-digit number is divisible by 4), we see that 54 is not divisible by 4 ($54/4 = 13.5$), but 84 is ($84/4 = 21$). Therefore, the answer is D.

19. (C) As in the problem above, the last two digits of the first 25 perfect squares are: $$ 01, 04, 09, 16, 25, 36, 49, 64, 81, 00, 21, 44, 69, 96, 25, 56, 89, 24, 61, 00, 41, 84, 29, 76, 25. $$ After $25^2$, the digits repeat symmetrically (e.g., $26^2$ ends like $24^2$). After $50^2$, the digits repeat normally (e.g., $51^2$ ends like $1^2$). So, looking at the list above, 41 is the only answer choice listed.