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Miscellaneous Number Theory: all problems

Focus mode: every practice problem from the chapter, one after another.

0 / 16 solved

Problem 28.3.1

AMC 12B

Suppose

2+11+12+23+x=14453.2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{2}{3 + x}}} = \frac{144}{53}.

What is the value of xx?

Problem 28.3.2

AMC 10

Driving along a highway, Megan noticed that her odometer showed 1595115951 (miles). This number is a palindrome: it reads the same forward and backward. Then 22 hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this 22-hour period?

Problem 28.3.3

AMC 10

A palindrome, such as 8343883438, is a number that remains the same when its digits are reversed. The numbers xx and x+32x + 32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of xx?

Problem 28.3.4

AMC 10

Mr. Zhou places all the integers from 11 to 225225 into a 1515 by 1515 grid. He places 11 in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?

Problem 28.3.5

AMC 10/12

A set of numbers is called sum-free if whenever xx and yy are (not necessarily distinct) elements of the set, x+yx + y is not an element of the set. For example, {1,4,6}\{1, 4, 6\} and the empty set are sum-free, but {2,4,5}\{2, 4, 5\} is not. What is the greatest possible number of elements in a sum-free subset of {1,2,3,…,20}\{1, 2, 3, \ldots, 20\}?

Problem 28.3.6

AMC 12Challenge

Let d(n)d(n) denote the number of positive integers that divide nn, including 11 and nn. For example, d(1)=1d(1) = 1, d(2)=2d(2) = 2, and d(12)=6d(12) = 6. (This function is known as the divisor function.) Let

f(n)=d(n)n3.f(n) = \frac{d(n)}{\sqrt[3]{n}}.

There is a unique positive integer NN such that f(N)>f(n)f(N) > f(n) for all positive integers n≠Nn \ne N. What is the sum of the digits of NN?

Problem 28.3.7(additional)

AMC 10/12

A palindrome between 10001000 and 10,00010{,}000 is chosen at random. What is the probability that it is divisible by 77?

4-digit palindromes: choose the first 2 digits, 9 × 10 = 90. 18 of the 90 4-digit palindromes are divisible by 7

digits4
divide by7

Tap a solid slot to change it. Dashed slots copy their mirror twin.

1,001 = 7 × 143: divisible by 7.

Divisible by 7: ? (hidden until the end of the example) of 90 4-digit palindromes

9 × 10 = 90 4-digit palindromes
9×101=909 \times 10^{1} = 90

Place value: abba‾=1001a+110b\overline{abba} = 1001a + 110b

1001 is a palindrome. 4-digit palindromes: 9 × 10 = 90 Divisible by 7: ? of 90 palindromes.

Problem 28.3.8(additional)

AMC 10

How many 77-digit palindromes (numbers that read the same backward as forward) can be formed using the digits 2,2,3,3,5,5,52, 2, 3, 3, 5, 5, 5?

Problem 28.3.9(additional)

AIME

Alice and Bob are playing a game with nn coins in a pile. Every turn, a player can take 11 or 44 coins from the pile, and Alice goes first. The player that takes the last coin wins. How many values of nn less than or equal to 20242024 are there such that no matter what Alice does, Bob has a winning strategy?

Problem 28.3.10(additional)

AMC 12

A palindrome is a nonnegative integer number that reads the same forwards and backwards when written in base 1010 with no leading zeros. A 66-digit palindrome nn is chosen uniformly at random. What is the probability that n11\dfrac{n}{11} is also a palindrome?

6-digit palindromes: choose the first 3 digits, 9 × 10 × 10 = 900. 900 of the 900 6-digit palindromes are divisible by 11

digits6
divide by11

Tap a solid slot to change it. Dashed slots copy their mirror twin.

123,321 = 11 × 11,211: divisible by 11.

Divisible by 11: 900 of 900 6-digit palindromes

9 × 10 × 10 = 900 6-digit palindromes
9×102=9009 \times 10^{2} = 900

Place value: abccba‾=100001a+10010b+1100c\overline{abccba} = 100001a + 10010b + 1100c

123321 is a palindrome. 6-digit palindromes: 9 × 10 × 10 = 900 Divisible by 11: 900 of 900 palindromes.

Problem 28.3.11(additional)

AMC 10

What is the ones digit of the sum

⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋?\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{2025} \rfloor?

(Recall that ⌊x⌋\lfloor x \rfloor represents the greatest integer less than or equal to xx.)

Problem 28.3.12(additional)

AMC 10

This problem also appears in the book as 21.3.18.

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is 11 year older than Chloe, and Zoe is exactly 11 year old today. Today is the first of the 99 birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

Problem 28.3.13(additional)

AIME

Ninety-four bricks, each measuring 4′′×10′′×19′′4'' \times 10'' \times 19'', are to be stacked one on top of another to form a tower 9494 bricks tall. Each brick can be oriented so it contributes 4′′4'' or 10′′10'' or 19′′19'' to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks?

Problem 28.3.14(additional)

AMC 10

Let NN be the positive integer 7777…7777777\ldots777, a 313313-digit number where each digit is a 77. Let f(r)f(r) be the leading digit of the rrth root of NN. What is

f(2)+f(3)+f(4)+f(5)+f(6)?f(2) + f(3) + f(4) + f(5) + f(6)?

Problem 28.3.15(additional)

AMC 10

An array of numbers is constructed beginning with the numbers −131-1 \quad 3 \quad 1 in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with −1-1 and 11, respectively.

−131−1241−11651\begin{array}{ccccccccc} & & -1 & & 3 & & 1 & & \\ & -1 & & 2 & & 4 & & 1 & \\ -1 & & 1 & & 6 & & 5 & & 1 \end{array}

If the process continues, one of the rows will sum to 12,28812{,}288. In that row, what is the third number from the left?

Problem 28.3.16(additional)

AIMEChallenge

Find the sum of all positive integers nn such that, given an unlimited supply of stamps of denominations 55, nn, and n+1n + 1 cents, 9191 cents is the greatest postage that cannot be formed.

Packs of 5, 24 and 25: the largest total you can't make is 91; 46 positive totals can't be made

pack a5
pack b24
pack c25

Predict first: which numbers can you make? Then fill in.

check a number11
11 can't be made from packs of 5, 24 and 25
Find the largest total you can't make.

Packs of 5, 24 and 25. 11 can't be made from packs of 5, 24 and 25. Find the largest total you can't make.