4 Mark Question

Question 14 Marks

Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9 .
(ii) The sum of three consecutive even numbers will be divisible by 6 .
(iii) If abcdef is a multiple of 6 , then badcef will be a multiple of 6 .
(iv) $8(7 b-3)-4(11 b+1)$ is a multiple of 12 .

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Question 24 Marks

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

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Question 34 Marks

The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

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Question 44 Marks

If $48 a 23 b$ is a multiple of 18 , list all possible pairs of values for $a$ and $b$.

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Question 54 Marks

Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9 ".
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9 ?

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Question 64 Marks

When is the sum of two multiples of 3 , a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

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Question 74 Marks

"I take a number that leaves a remainder of 8 when divided by $12 . I$ take another number which is 4 short of a multiple of 12 . Their sum will always be a multiple of 8 ", claims Snehal. Examine his claim and justify your conclusion.

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Question 84 Marks

If $31 z 5$ is a multiple of 9 , where $z$ is a digit, what is the value of $z$ ? Explain why there are two answers to this problem.

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Question 94 Marks

"I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up - it simply won't do, A stubborn odd pebble remains in my view. Group them by 5 , yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?"

Answer

Grouped in 3's leaves 1.
Pairing (2's) leaves 1.
Grouped by 5 leaves 1.
Grouped by 7 is perfect.
Number $\leq 100$.
L.C.M of 2,3 , and $5=30$.
In all those cases, when we group them, 1 pebble remains.
So, the actual number of pebbles must be $=30+1=31$, but 31 is not divisible by 7 .
The next multiple of 30 is $2 \times 30=60$.
So, $60+1=61$, but this is also not divisible by 7 .
Similarly, the next number is $90+1=91$.
And 91 is divisible by 7 .
Hence, the number of pebbles I hold $=91$.

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