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AQA GCSE · Question 09 · Number
When x is divided by 2 the remainder is 1.
When x is divided by 3 the remainder is 1.
When x is divided by 4 the remainder is 1.
Work out two possible values of x.
When x is divided by 2 the remainder is 1.
When x is divided by 3 the remainder is 1.
When x is divided by 4 the remainder is 1.
Work out two possible values of x.
How to approach this question
1. Interpret the information. If dividing x by a number gives a remainder of 1, it means x is one more than a multiple of that number.
2. This means that (x - 1) must be a multiple of 2, 3, and 4.
3. Find the Lowest Common Multiple (LCM) of 2, 3, and 4.
4. The value of (x - 1) will be the LCM, or any multiple of the LCM.
5. Find the first multiple of the LCM, add 1 to get the first value of x.
6. Find the second multiple of the LCM, add 1 to get the second value of x.
Full Answer
The problem states that when x is divided by 2, 3, or 4, the remainder is always 1.
This can be expressed algebraically:
x = 2a + 1
x = 3b + 1
x = 4c + 1
(where a, b, c are integers)
If we subtract 1 from x in each case, we get:
x - 1 = 2a
x - 1 = 3b
x - 1 = 4c
This shows that the number (x - 1) is a common multiple of 2, 3, and 4.
To find the possible values of x, we first find the Lowest Common Multiple (LCM) of 2, 3, and 4.
The multiples of 4 are 4, 8, 12, 16...
12 is also a multiple of 2 and 3. So, LCM(2, 3, 4) = 12.
This means (x - 1) must be a multiple of 12.
The multiples of 12 are 12, 24, 36, 48, ...
So, we can have:
x - 1 = 12 => x = 13
x - 1 = 24 => x = 25
x - 1 = 36 => x = 37
...and so on.
The question asks for two possible values, so we can choose any two from this list, for example, 13 and 25.
Common mistakes
✗ Finding the product of 2, 3, and 4 (which is 24) instead of the LCM (12). This would give correct values (25, 49) but misses the smallest possible value (13).
✗ Forgetting to add 1 back at the end.
✗ Listing multiples of 12 (12, 24) as the answer.
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