Number Play: questions with answers

The sum $3 + 5 + 7 + 8 = 23$ is odd. Switching a sign changes the result by twice the corresponding number, which is even. Adding or subtracting an even number keeps an odd result odd. Therefore, all eight results are odd, including the negative results.

Switching the sign before a number changes the value by twice that number. The two chains differ only in the signs before 2, 3 and 5. So the whole change is 2 × (2 + 3 + 5) = 20, which is even. An even change keeps the parity, so 11 and -9 have the same parity. The number of minus signs does not matter, since each switch moves the value by an even amount. A negative multiple of 2 is even in the same way a positive one is.

The expressions generate exactly the same numbers. Using k + 1 in the minus form gives $6(k + 1) - 4 = 6k + 2$. Using k - 1 in the plus form gives $6(k - 1) + 2 = 6k - 4$. These inputs are integers, so every number generated by either expression also occurs in the other.

No. For k = 1, 2, 3, 3k + 5 gives 8, 11, 14, and 11 leaves remainder 1 when divided by 5. The numbers that leave remainder 3 are 3, 8, 13, …, given by 5k + 3: the divisor 5 multiplies k and the remainder 3 is added.

Squaring preserves parity: an even integer has an even square, and an odd integer has an odd square. Two integers have an even sum exactly when they have the same parity. Since 10 is even, $x^2 + 10$ is even exactly when $x$ is even.

The product is even. The factor 10 is a multiple of 2, so the whole product is a multiple of 2. The odd factors 809 and 213 do not change this. Changing the order of multiplication leaves the product unchanged.

Adding gives $4u + 4v + 4$, since $2 + 2 = 4$. Taking out the common factor of 4 gives $4(u + v + 1)$. This is a multiple of 4 because $u + v + 1$ is a whole number.

The four sums are $16$, $22$, $18$, and $24$. Exactly two, $16$ and $24$, are multiples of 4. Two even numbers give a multiple of 4 when their remainders on division by 4 are equal. The remainders can be 0 and 0, or 2 and 2. Remainders of 0 and 2 give a sum with remainder 2.

No -- 10 is divisible by 5 and is not divisible by 40. A single factor test can rule a number OUT for certain: failing the test for 5 means 40 is impossible, since 5 divides 40. What it cannot do is rule one in, because it says nothing about the part of 40 that 8 is responsible for. So the answer is not that nothing follows -- a great deal follows in the failing direction -- and it is not that the pair multiplying to 40 settles it either, since both tests still have to be passed.

Write the two numbers as 6a and 6b. Their sum is 6(a + b) and their difference is 6(a - b). Both carry the factor 6, so 6 divides 72 and 12. Taking whole groups away from whole groups leaves whole groups. The rule does not run backwards. For example, 32 + 40 = 72, but neither 32 nor 40 is a multiple of 6.

For P = 1, Q = 3, R = 7, S = 8, the multiplication is $13 \times 6 = 78$. All four letters stand for different digits, and each number begins with a nonzero digit. The other assignments either repeat a digit for different letters or put 0 at the start of PQ.

Yes, and it is certain rather than lucky. The test for 9 reads only the SUM of the digits, and reversing rearranges the digits without adding or removing any, so the sum is still 18 -- a multiple of 9 -- and so is 954. This is why Sreelatha is right about every multiple of 9, not just the ones she tried, and why any other shuffle of the digits works too. The rule runs in both directions: a digit sum divisible by 9 does not merely suggest that 954 is a multiple of 9, it settles it, so no division is needed to confirm the answer.

For divisibility by 4, check the last two digits. For 532, these form 32, and 32 = 4 × 8. So 532 is divisible by 4. For 530, the last two digits form 30, which is not a multiple of 4. Its units digit alone gives the wrong answer. The remaining option, 531, is odd, so it cannot be divisible by 4.

4. The alternating sum of 7025 is -4, so 7025 is 4 short of a multiple of 11 -- which is exactly the amount that has to be added. The number 7 answers the other half of the same fact: it is how far 7025 sits ABOVE the multiple below it, and the two add to 11. Both describe the same position and only one of them was asked for. The value -4 itself is the signed shortfall and is not an amount to add.

The digital root is 4. Adding the digits of 58396 gives 31. Adding the digits of 31 gives 4. Repeat if needed until one digit remains: 4. Stopping at 31 is too soon because it still has two digits. The units digit 6 alone does not give the digital root.