Consider the statement: ‘If a number is divisible by 8, then it is divisible by both 2 and 4’.
- (i) Justify the statement.
- (ii) Recall the divisibility shortcuts that we have studied for different numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisible by 2 and 4? Why or why not?
Step-by-step solution
Idea: The statement is true, but checking 2 and 4 is checking its converse, which is false. Since 4 already contains the factor 2, ‘divisible by 2 and 4’ is no more than ‘divisible by 4’.
(i) Justify the statement.
- A number divisible by 8 is 8k for some whole number k.½ mark
- 8k = 2 × 4k and 8k = 4 × 2k, so it is divisible by both 2 and 4. The statement is true.½ mark
(ii) Recall the divisibility shortcuts that we have studied for different numbers. To check whether a given number is divisible by 8, is it enough to check whether it is divisible by 2 and 4? Why or why not?
- No. Using 2 and 4 to test for 8 relies on the converse: ‘If a number is divisible by both 2 and 4, then it is divisible by 8’.½ mark
- The converse is false. 12 = 2 × 6 = 4 × 3 is divisible by 2 and by 4, but 12 = 8 × 1 + 4 is not divisible by 8. (20, 28 and 36 also fail.)½ mark
- Why: every multiple of 4 is already even, so ‘divisible by 2 and 4’ just means ‘divisible by 4’. It does not supply the extra factor 2 needed for 8 = 2 × 2 × 2.½ mark
- The right shortcut for 8: check whether the number formed by the last three digits is divisible by 8. E.g. 3124 is divisible by 4 (24 ÷ 4 = 6), but 124 = 8 × 15 + 4, so 3124 is not divisible by 8.½ mark
Check: Multiples of 4 up to 40: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 — only every second one (8, 16, 24, 32, 40) is a multiple of 8.
Answer to write in the exam
(i)
Let the number be 8k.
8k = 2 × 4k = 4 × 2k
∴ It is divisible by both 2 and 4.
(ii)
No.
Converse: If a number is divisible by both 2 and 4, then it is divisible by 8 — false.
Counterexample: 12 = 2 × 6 = 4 × 3, but 12 = 8 × 1 + 4.
Divisible by 4 ⇒ divisible by 2, so ‘2 and 4’ gives only divisibility by 4.
∴ Not enough; check whether the last three digits form a multiple of 8.
Common mistakes that cost marks
- Thinking ‘divisible by 2 and 4’ means ‘divisible by 2 × 4 = 8’. That multiplication rule works only for numbers with no common factor, and 2 and 4 share 2.
- Testing only the last two digits for 8. Last two digits test divisibility by 4; for 8 use the last three digits.
- Justifying (i) with one example such as 16 instead of the general 8k.
How this can come in the exam
Which number is divisible by 4 but NOT by 8?
- 1,000
- 2,036
- 3,016
- 5,120
Show answer
(B) 2,036
2036 = 4 × 509, but 036 = 36 = 8 × 4 + 4, so 2036 is not divisible by 8. 1000, 3016 and 5120 are multiples of 8.
Assertion (A): A number divisible by both 3 and 4 is divisible by 12.
Reason (R): A number divisible by both 2 and 4 is divisible by 8.
- Both Assertion (A) and Reason (R) are true and R is the correct explanation of A.
- Both Assertion (A) and Reason (R) are true but R is not the correct explanation of A.
- Assertion (A) is true but Reason (R) is false.
- Assertion (A) is false but Reason (R) is true.
Show answer
(C) Assertion (A) is true but Reason (R) is false.
A is true because 3 and 4 have no common factor other than 1. R is false: 12 is divisible by 2 and 4 but not by 8.
Try one yourself
Is checking ‘divisible by 2 and by 9’ enough to decide whether a number is divisible by 18? Why?
Show answer
Yes. 2 and 9 have no common factor other than 1, so a number divisible by both is divisible by 2 × 9 = 18 (unlike 2 and 4 for 8).
More questions like this
- Recall that a shortcut to check whether a given number is divisible by 3 is to add the digits of the number and check if the sum is a multiple of 3. Express the relationship between ‘a number is divisible by 3’ and ‘sum of the digits is a multiple of 3’ using ‘If-then’ sentences.
- We have identified different types of quadrilaterals — squares, rectangles, parallelograms, rhombi, kites and trapezia. One can identify more types (e.g., we could create a category of quadrilaterals that have equal-length opposite sides).
Suppose we have identified a category of quadrilaterals called Q, and we have to construct a quadrilateral of this type. For this, we are to use two thin sticks, put them together as diagonals so that the quadrilateral obtained by joining their endpoints is of type Q (see the figure). - If a proposition is true, then is its converse always true?
- Statement 1: If two sides of a triangle are equal, then the angles opposite the equal sides are equal.
Statement 2: If two angles of a triangle are equal, then the sides opposite the equal angles have equal lengths.
We have proved the first statement in an earlier grade. Is the second statement true? Can you prove it?
(Hint: Draw the altitude from the vertex containing the third angle.) - Proposition P: If it rains, then the road is wet.
Converse Q: If the road is wet, then it has rained.