If t is a positive integer, can (t+2)(t-3) be completely divided by 6?

Question

(1)  t^2  +1 is not divisible by 2.

(2) t is a 3-digit number, whose digits are consecutive integers.

chetan2u · Answer

(t+2)(t-3)=t^2-t-6  now  6 will be divisible by 6 , so we have to see whether  t^2-t=t(t-1)  is divisible by 6.
 t and t-1 are consecutive integers, so their product would surely be divisible by 2.  Thus, we have to check whether any of t or t-1 is multiple of 3 ..

(1)  t^2  +1 is not divisible by 2.
So, t is even..
If t is 6, ans is yes as t(t-1) or 6*5 will be divisible by 3.
However, if t is 2, then 2(2-1) or 2 is NOT divisible by 3.
Insuff

(2) t is a 3-digit number, whose digits are consecutive integers.
Let the smallest digit be a, so other two digits will be a+1 and a+2..
For a number to be divisible by 3, sum of its digits must be divisible by 3..
Let us check the sum : a+(a+1)+(a+2)=3a+3=3(a+1). Hence divisible by 3.
Answer is YES
Suff

B

Sneha2021 · Answer

What's the meaning of "whose digits are consecutive integers"?

St2 - T can be 123,234,456 or 123,124,125?

Can someone please help?

DamanDeep · Answer

Answer is B

As per statement 2.
t is 3 digit number with consecutive digits.

t = n(n+1)(n+2) ; sum of digits : 3n+3 = 3(n+1); so t is always divisible by 3. 
Hence we can say (t+3) from equation is always divisible by 3.

Next is to check for 2;
case 1 : if t is odd; t-3 is divisible by 2.
case 2 : if t is even; t+2 is divisible by 2.
in both the cases, there is win win situation.

Hope u got it.

bumpbot · Answer

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If t is a positive integer, can (t+2)(t-3) be completely divided by 6?

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