a. TRUE
Let the consecutive numbers be (n - 1), n and (n + 1)
Their sum: (n - 1) + n + (n + 1) = 3n which is always a multiple of 3.
For e.g. 1 + 2 + 3 = 6 = 3*2
14 + 15 + 16 = 45 = 3*15

b. TRUE
Let the numbers be (2n - 3), 2n - 1, (2n + 1), (2n + 3)
Their sum: 8n, which is a multiple of 8.
For e.g. 1 + 3 + 5 + 7 = 16 = 8*2

c. TRUE (assuming counting = consecutive)
(a^2 - b^2) = (a - b)(a + b)
Since a,b are both odd, let a = 2x +1 and b = 2x - 1.
So, (a^2 - b^2) = (4x)(2) = 8 xy, and is a multiple of 8.
For e.g., when a = 7, and b = 5, (a^2 - b^2) = 24, A MULTIPLE of 4.


d. TRUE
n(n + 1)(n + 2)(n + 3)
Since the numbers are consecutive, two of them are even, one of time is a multiple of 4 and one of them is a multiple of 3.
n(n + 1)(n + 2)(n + 3) = (2x)(4y)(3z)(t) = 24 xyzt
So, this product is always a multiple of 24.
For e.g. (1)(2)(3)(4) = 24*1
(4)(5)(6)(7) = 840 = 24*35

e. NOT IN ALL CASES
a = b^2 (after squaring both sides)
if b = 1/2, a = 1/4. b > a
if b = 2, a = 4. a > b
So, a > b only when b > 1.

hannah_lewis11 wrote:For each of the following statements, state whether it is true or false. If a statement is true, give at least 2 examples that demonstrates that it is true. If a statement us false, give 1 example that shows it is false

a, The sum of three consecutive counting numbers is always a multiple of 3.

b, The sum of four consecutive odd numbers is a multiple of 8.

c, If a and b are odd counting numbers with a > b, then a^2 - b^2 is a multiple of 8

d, The product of four consecutive counting numbers is a multiple of 24

e, If a is a positive number and √a=b, then a > b

If anyone could answer any of these questions it would be much appreciated cheers

a. If sum of digits of a no. is divisible by 3, then no. is divisible by 3. True.

b. say consecutive odd nos. are 2x+1, 2x+3, 2x+5 & 2x+7. their sum is 8x+16 = 8.(x+2). True.

c.Say the odd nos. a= 2x+3 and b= 2x+1.

a^2 - b^2 = (a+b).(a-b) = (4x+4).2 = 8.(x+1). True.

You may try with other sets also. Say a= 2x+11 and b= 2x+1.

a^2 - b^2 = (a+b).(a-b) = (4x+12).10 = 40.(x+3). True.

d. We can observe that multiple of 2 occurs in every second alternative no., multiple of 3 occurs in every third no. & multiple of 4 occurs in every 4th no.

So for any set of consecutive 4 nos. At least one multiple of 2, 3 & 4 will be present. This makes 2.3.4 = 24. True.

e.False. If a = 4, b = 2. true, but if a=1/4, then b=1/2. false.

d. TRUE
n(n + 1)(n + 2)(n + 3)
Since the numbers are consecutive, two of them are even, one of time is a multiple of 4 and one of them is a multiple of 3.
n(n + 1)(n + 2)(n + 3) = (2x)(4y)(3z)(t) = 24 xyzt
So, this product is always a multiple of 24.
For e.g. (1)(2)(3)(4) = 24*1
(4)(5)(6)(7) = 840 = 24*35


All the other options turned out to be correct for me, but the option D knocked me out...


try as many values as possible, but the answer turns out to be true.

I multiplied n (n+1) (n+2) (n+3) and got the final answer but I do not think that really helped me solve the question.

The explanation given by aneesh, did not strike to me. SO my question is if you go by the formula then what should be the answer.

You could put the value in the above equation anything greater than 0 i.e in the equation n (n+1) (n+2) (n+3).

Not 0 as the product is asked. You will find that 24 or its multiple is definitely present.

I hope this post was helpful...

For each of the following statements, state whether it is true or false. If a statement is true, give at least 2 examples that demonstrates that it is true. If a statement us false, give 1 example that shows it is false

a, The sum of three consecutive counting numbers is always a multiple of 3.

b, The sum of four consecutive odd numbers is a multiple of 8.

c, If a and b are odd counting numbers with a > b, then a^2 - b^2 is a multiple of 8

d, The product of four consecutive counting numbers is a multiple of 24

e, If a is a positive number and √a=b, then a > b

For all these problems, the best way forward is to assign values to a, b, c and then solve.

a) Let the three numbers be n, n+1 and n+2. Sum = n+(n+1)+(n+2) = 3n+6 = 3(n+2), a multiple of 3, hence a is true

b) Let the odd numbers be 2n+1, 2n+3, 2n+5 and 2n+7. Their sum = 8n+16 = 8(n+2), a multiple of 8,
hence b is true

c) let a= 2n+1 and b=2n-1. so, a^2 - b^2 = (a-b)(a+b) = {(2n+1)+(2n-1)} {(2n+1)-(2n-1)}=4nX2 =8n, a multiple of 8. Hence, c is true

d) Let the numbers be n, n+1, n+2, n+3. Their product = n(n+1)(n+2)(n+3). Assign values to n and see.
Put n=1, product is 24, hence true, assign n=3, product is 3*4*5*6 = 360, divisible by 24. In this way, this will hold for all values of n. Hence, d is true

e) This is true for all values of a>1, for example for a=4, b=2, for a=9, b=3, etc.
However, for a=1/4, b=1/2, b>a. And for a=1, b=1, => a=b. Hence, this may or may not be true. Hence, e is true

Remember the main takeaway from these problems is to remember that these are properties of numbers which can be used for solving many problems. There will be many problems where you will need to apply these properties for quick problem-solving!