Quiz

Hint

The pattern repeats every 12 hours. So after 12, 24, 36, ……..and 996 hours it will be 5 o’clock. In 1000 hours it will be 4 hours later which will be 9 o’clock.

Hint

All we need to do is add multiples of 5 to the smallest integer greater than 5 which is congruent to 3 modulo \(5: 8,13,18,23\), and 28 are all congruent to \(3(\bmod 5)\). We can see that the remainder of each of the numbers above when divided by 5 is 3.

Hint

We can see that 3,300 is just \(3(1,100)\) and creates three remainders of 8, or 24: If \(1,100 \equiv 8(\bmod 12)\), then \(3(1,100) \equiv 3 \cdot 8(\bmod 12) \equiv 24(\bmod 12) \equiv 0(\bmod 12)\).

Hint

There are 30 days in June, for a total of 60 twelve-hour periods. \(19 \equiv 7(\bmod 12)\) and \(60 \equiv 0(\bmod 12)\). We can see then that \(19 \cdot 60 \equiv 7 \cdot 0(\bmod 12) \equiv 0(\bmod 12)\). At the beginning of July, Parker’s clock will be correct and will read 6 o’clock again.

Hint

\(
\begin{aligned}
& 3^{1}=3(\bmod 8) \\
& 3^2=3 \cdot 3=9 \equiv 1(\bmod 8) \\
& 3^{3}=3 \cdot 3^2 \equiv 3 \cdot 1(\bmod 8) \equiv 3(\bmod 8) \\
& 3^5=3^2 \cdot 3^3 \equiv 1 \cdot 3(\bmod 8) \equiv 3(\bmod 8)
\end{aligned}
\)
We can continue in this way to show that every odd power of 3 is congruent to \(3(\bmod 8)\). Adding the first 15 odd powers of 3 gives us:
\(
3^{\prime}+3^3+3^s+\ldots+3^s \equiv 3 \cdot 15(\bmod 8) \equiv 5(\bmod 8) \text {. }
\)
The remainder is therefore 5 .

Hint

PRIYANKA has 8 letters. 955 is 3(mod8), so she completes her name and writes 3 more letters. The third letter in her name is I.

Hint

\((1+2+3+\ldots+48)\) is equal to \(\frac{(48)(49)}{2}=24(49)\) \(49 \equiv-1(\bmod 50)\), so \(24(49) \equiv-24(\bmod 50)\), and \(-24(\bmod 50) \equiv 26(\bmod 50)\). We are looking for the value of \(26^{49}\) in modulo \(50.26^1 \equiv 26(\bmod 50)\) and \(26^2=676 \equiv 26(\bmod 50)\), so \(26^{49} \equiv 26(\bmod 50)\) (every power of 26 ends in 76). \((1+2+\ldots+48)^{49}\) leaves a remainder of 26 when divided by 50 .

Hint

If we call Katy student 1, then Katy \(\equiv 1(\bmod 8)\), and Meera wants to be \(\equiv 1(\bmod 8)\). \(350 \equiv 6(\bmod 8)\), and counting backwards we get \(345 \equiv 1(\bmod 8)\), which is 6 places to Katy’s right (there are 5 students between Meera and Katy).

Hint

We are looking for \(n\) such that \(31 n \equiv 1(\bmod 13)\), but 31 is congruent to 5 in modulo 13. This simplifies our expression to \(5 n \equiv 1(\bmod 13)\). We see that this works when \(n=8\), which gives us \(40 \equiv 1(\bmod 13)\). The eighth multiple of 31 will therefore leave a remainder of 1 when divided by \(13: 31 \cdot \mathbf{8}=\mathbf{2 4 8}\).

Hint

We are trying to advance the clock ahead by one minute. If we push the \(>\) button 9 times, we increase the time on the clock by 3 minutes: \(9 \cdot 7=63 \equiv 3(\bmod 60)\). If we made 5 “laps” using this method (by pushing \(>45\) times) we would end up 15 minutes ahead of where we started:
\(5 \cdot 3=15 \equiv 1(\bmod 7)\). Pushing \(>2\) less times would put us one minute ahead at 6:04 (in 43 button presses).
However, if we push \(<8\) times, we end up 4 minutes ahead: \(8(-7)=-56 \equiv 4(\bmod 60)\). Taking two “laps” in this manner by pressing \(<16\) times, we are left 8 minutes ahead: \(2 \cdot 4=8 \equiv 1(\bmod 7)\). One more press of \(<\) and we are at 6:04, one minute ahead of where we started in a total of \(\mathbf{1 7}\) presses of \(<\).

Hint

We begin by finding an integer \(x\) where \(x \equiv 1(\bmod 14)\) and \(x \equiv 2(\bmod 15)\). Looking for the smallest positive integer which leaves a remainder of 1 when divided by 14 and 2 when divided by 15 can be difficult, but notice that the difference between 14 and 15 is equal to the difference in the modular residues 1 and 2 . This leads us to look for the first negative solution \(x=-13\). Add to this \(14 \cdot 15=210\) to maintain modular congruence and we get 197 trees. Unfortunately, the problem states that Craig has more than 200 trees so we must add another 210 to get 407 trees.

Hint

Look for a pattern: \(2^0\) leaves a remainder of \(1,2^0+2^1\) leaves a remainder of \(3,2^0+2^1+2^2\) leaves a remainder of 7, and the pattern of remainders continues: \(1-3-7-6-4-0-1-3-7-6-4\). \(0 \ldots\) in a repeating block of 6 digits. 99 leaves a remainder of 3 when divided by 6 , so the remainder will be the same as for \(2^0+2^1+2^2+2^3\), which is 6.

Hint

\(199 \text { is prime. The product of its integer factors is} 199\)

Hint

The prime factorization of 900 is \(2^2 \cdot 3^2 \cdot 5^2\). We can remove a 2 , a 3 , or a 5 and leave a number that has 18 factors, so there are 3 factors of 900 which have exactly 18 factors \((180,300\), and 450\()\)

Hint

The sum of the five digits must be 9, and the last digit must be a 2 (a number divisible by 8 must be even). We want to place as many leading ones as possible to keep the integer as small as possible while keeping the sum of the digits equal to 9 and using at least one 3: this leads us to \(\mathbf{1 1, 2 3 2}\), which is divisible by both 8 and 9.

Hint

We are looking for the remainder when 500 is divided by 6, which is 2 (the \(498^{\text {th }}\) person will count off 6, with the last two people numbered 1 and 2 ).

Hint

First, we find the smallest integer that is divisible by 2, \(3,4,5,6,8,9,10\), and \(11: 2^3 \cdot 3^2 \cdot 5 \cdot 11=3,960\). Any multiple of 3,960 is divisible by \(2,3,4,5,6,8,9,10\), and 11. There are 2 multiples of 3,960 less than \(10,000: 3,960\) and 7,920.

Hint

There are an enormous number of sets of positive integers whose sum is 25, so there must be a method. Let’s look for a pattern. For 6 : \(1+5=6\) and \(1 \cdot 5=5\), \(2+2+2=6\) and \(2 \cdot 2 \cdot 2=8\), \(3+3=6\) and \(3 \cdot 3=9\).
It appears that we are better off using 3 ‘s (not 2 ‘s) if we want to maximize the product for a given sum. Let’s compare 3’s to 5’s: \(3+3+3+3+3=15\) and \(3^5=243\) while \(5+5+5=15\) and \(5^3=125\). 3 ‘s are better than 5 ‘s when trying to achieve a given sum while maximizing a product for a set of integers. 3 ‘s clearly do better than the larger primes like 7 and 11. We, therefore, look to use as many 3’s as possible, with any remainder used in 2’s. If the remainder is 1 then we are better off using one less three so that we leave a remainder of \(4=2 \cdot 2\), which exceeds \(1 \cdot 3\). For a sum of 25 we can use seven 3 ‘s and two 2’s: \(3(7)+2(2)=25,3^7 \cdot 2^2=\mathbf{8}, 748\)

Hint

We are looking to find a value \(x\) where \(38 x \equiv-1(\bmod 35)\). Because \(38 \equiv 3(\bmod 35)\), we simplify our search to \(3 x \equiv-1(\bmod 35)\). Note that any number which is congruent to \(-1(\bmod 70)\) is also congruent to \(-1(\bmod 35)\), which helps us see that \(x=23\) is a solution. 23 students in 38 groups = 874 students ( 25 rows of \(35=875\) seats).

Hint

The tires have circumference \(27 \pi\) inches and \(30 \pi\) inches. The LCM \((270 \pi)\) inches gives us the distance after which the tires will both reach their original position in the rotation. The stars will be positioned as shown in the drawing 5 times (every \(54 \pi\) inches) during this period. \(1 / 4\) mile \(=15,840\) inches. \(15,840 /(54 \pi)=93.4\).
If the wheels begin in the position shown, this makes 94 times.

Hint

For any pair of consecutive integers, one must be even and therefore divisible by 2 , so we look for the least positive integer whose prime factorization is of the form \(2^2 \cdot a\) or \(a^2 \cdot 2\) for which an adjacent integer has 6 factors. \(2^2 \cdot 11=44\) and \(3^2 \cdot 5=45\) are the first pair of consecutive integers which have exactly 6 factors each. Pairs like \((75,76)\) and \((98,99)\) work as well.

Hint

\(
7^{100} \equiv\left(7^2\right)^{50} \equiv 49^{50} \equiv(-1)^{50} \equiv 1 \bmod 10
\)

Hint

Note that \(3^4 \equiv 81 \equiv 1 \bmod 5\). Let \({x}\) be defined as \(x \equiv s \bmod 20\), where \(s \leq 20\). Then \(x \equiv s \bmod 4\) and \(x \equiv s \bmod 5\). These imply that \(3^x \equiv 3^s \bmod 20\) and \(x^2 \equiv s^2 \bmod 20\), so \(3^x-x^2 \equiv 3^s-s^2 \equiv 0 \bmod 20\)

After trying values, you will find that \(s=2,4,16\), or 18 are the only values possible. Thus, that are \(4 \times 5=20\) possible values of \(x \leq 100\).

Hint

\(30 a 0 b 03=3,000,003+a * 10,000+b * 100 \equiv 400,003+a * 900+b * 9 \equiv 10,003+a *(-10)+b * 9 \equiv\)
\(903+a * 3+b * 9 \equiv(-7)+a * 3+b * 9 \equiv 3 a+9 b-7 \equiv 0 \bmod 13\).
Therefore, \(3 a+9 b \equiv 7 \equiv 20 \equiv 33\) \(\bmod 13\).
After dividing by 3 because \((3,13)=1\), we get \(a+3 b \equiv 11 \bmod 13\).
Thus we have \(a=2, b=3\) or \(a=5, b=2\) or \(a=8, b=1\).

Hint

First of all, we know that \(k ! \equiv 0(\bmod 9)\) for all \(k \geq 6\). Thus, we only need to find \((1 !+2 !+3 !+4 !+5 !)(\bmod 9)\)
\(
\begin{aligned}
& 1 ! \equiv 1(\bmod 9) \\
& 2 ! \equiv 2(\bmod 9) \\
& 3 ! \equiv 6(\bmod 9) \\
& 4 ! \equiv 24 \equiv 6(\bmod 9) \\
& 5 ! \equiv 5 \cdot 6 \equiv 30 \equiv 3(\bmod 9) \\
&
\end{aligned}
\)
Thus,
\(
(1 !+2 !+3 !+4 !+5 !) \equiv 1+2+6+6+3 \equiv 18 \equiv 0(\bmod 9)
\)
so the remainder is 0.

Hint

\((n+1)(n+2) \equiv 0 \equiv 2 \times 3 \equiv 5 \times 6 \equiv 6 \times 1 \bmod 6\). Therefore \((n+1) \equiv 2,5,6 \bmod 6\), and \(n \equiv 1,4,5 \bmod 6\). There are \(5+4+4=13\) of these numbers between 1 and 25.

Hint

The sum of the first three numbers is divisible by 3 . The sum of the first four numbers is divisible by 4 . If we write out all 5 numbers in \(\bmod 3\), we get \(2,1,2,1,1\), respectively. Clearly, the only way to get a number divisible by 3 by adding three of these is \(1+1+1\), so those scores must be entered first. Now we have an odd sum, so we must add 71 in order for the sum to be divisible by 4. That leaves 80 for the last score entered.

Hint

There are either \(65+200=265\) or \(66+200=266\) days between the first two dates depending upon whether or not year \(N\) is a leap year. Since 7 divides into 266, then it is possible for both dates to Tuesday; hence year \(N+1\) is a leap year and \(N-1\) is not a leap year. There are \(265+300=565\) days between the date in years \(N, N-1\), which leaves a remainder of 5 upon division by 7 . Since we are subtracting days, we count 5 days before Tuesday, which gives us Thursday.

Hint

\(
71-x=8 n
\)
\(71-x\) is the multiple of 8
if \(x=1,71-x=70\) is not the multiple of 8 ,
The number which is nearest to 71 is 64 (multiple of 8). In order to make this 71 as 64 , we have to subtract 7 .
So, the value of \(x\) is 7
Hence 7 is the least number.