What is the remainder when \(x^2 - 7x + 15\) is divided by \(x-3\)?
Answer: A
By the Polynomial Remainder Theorem, the remainder when a polynomial P(x) is divided by \(x-a\) is P(a). Here, P(x) = \(x^2 - 7x + 15\) and a=3. We calculate P(3). \(P(3) = (3)^2 - 7(3) + 15 = 9 - 21 + 15 = 24 - 21 = 3\). The remainder is 3.
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A 4-digit number is formed by repeating a 2-digit number such as 2525 or 3232. Any number of this form is always exactly divisible by:
Answer: D
Let the 2-digit number be 'ab'. The 4-digit number formed is 'abab'. This can be written mathematically as \(1000a + 100b + 10a + b = 1010a + 101b = 101(10a+b)\). Since \(10a+b\) represents the original 2-digit number, the 4-digit number 'abab' is always a multiple of 101. Therefore, it is always exactly divisible by 101.
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What is the remainder when \(397^{3589} + 5\) is divided by 398?
Answer: A
We can write 397 as \(398-1\). The expression becomes \((398-1)^{3589} + 5 \pmod{398}\). The first term \((398-1)^{3589}\) has a remainder of \((-1)^{3589} = -1\) when divided by 398. The second term, 5, has a remainder of 5. The total remainder is \(-1 + 5 = 4\). The remainder is 4.
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The number 23473 is to be divided by 7. What is the remainder?
Answer: B
Correct answer: 2
The doubling rule for 7 tests divisibility but does not hand you the remainder, so divide directly.
\[23473 = 7\times 3353 + 2\]
Since \(7\times 3353 = 23471\), the remainder is \(23473 - 23471 = 2\).
As a check, \(23471\) is exactly divisible by 7, and 23473 sits two above it.
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The number 23473 is to be divided by 7. What is the remainder?
Answer: B
To find the remainder, we can perform long division.
\(23 \div 7 = 3\) with a remainder of 2. Carry over 2.
\(24 \div 7 = 3\) with a remainder of 3. Carry over 3.
\(37 \div 7 = 5\) with a remainder of 2. Carry over 2.
\(23 \div 7 = 3\) with a remainder of 2.
The final remainder is 2. Alternatively, \(23473 = 21000 + 2100 + 350 + 21 + 2\). All terms except the last one are divisible by 7. So the remainder is 2.
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What is the smallest number that must be added to 1056 to make the sum completely divisible by 23?
Answer: A
First, we divide 1056 by 23 to find the remainder. \(1056 = 45 \times 23 + 21\). The remainder is 21. The number that must be added to make the sum divisible by 23 is \(\text{Divisor} - \text{Remainder} = 23 - 21 = 2\).
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What is the remainder when \(17^{23}\) is divided by 16?
Answer: A
We can express the base 17 as \(16+1\). So, we need to find the remainder of \((16+1)^{23} \div 16\). According to the remainder theorem, this is equivalent to finding the remainder of \(1^{23} \div 16\). Since \(1^{23} = 1\), the remainder is 1.
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A number is divisible by 13. If we divide that number by 5, it leaves a remainder of 3. What is the smallest such number?
Answer: D
Correct answer: 13
List the multiples of 13 in order and test each against the second condition.
\[13 \div 5 = 2 \text{ remainder } 3\]
The very first multiple already satisfies it, so 13 is the smallest such number. For comparison, 26 leaves remainder 1, 39 leaves 4 and 52 leaves 2, so none of the larger options qualifies as the smallest.
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What is the smallest number that must be added to 1057 to make it exactly divisible by 23 ?
Answer: A
Correct answer: 1
\[1057 = 23 \times 45 + 22\]
The remainder is 22, so the next multiple of 23 is one step away.
\[23 - 22 = 1, \qquad 1057 + 1 = 1058 = 23 \times 46\]
Note the contrast with subtraction, where the remainder itself is removed.
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Find the remainder when \(3^{81}\) is divided by 82.
Answer: D
Correct answer: 3
Here \(82 = 2\times 41\) and 3 shares no factor with either, so Euler’s theorem applies.
\[\phi(82) = \phi(2)\,\phi(41) = 1\times 40 = 40 \quad\Longrightarrow\quad 3^{40} \equiv 1 \pmod{82}\]
Since \(81 = 2\times 40 + 1\),
\[3^{81} = \left(3^{40}\right)^{2}\times 3 \equiv 1^{2}\times 3 = 3 \pmod{82}\]
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