Question

Today it is my birthday. My age in
years has 8 factors (including 1
and itself). Last year it only had 4
factors. Next year, it will only have
two. This sequence of 4, 8, 2
factors has never happened to me
before. How old am I, and when
will this sequence of factors
happen again?
​

Answer 1

I don't know

sorry π_π π_π

〒_〒

Answer 2

Solution :-

• Present age = 8 factors . => (1 + 1)(1 + 1)(1 + 1) => multiple of 3 prime numbers or, (3 +1)(1+1) = a³*b .
• Last year = 4 factors . => (1 + 1)(1 + 1) => Multiple of 2 prime numbers .
• Next year = only two . => A prime number .

check :-

• next year ≠ 2, 3, 5, 7 , 11, 13, 17 , 19 ,23, 29, 31 , 37 .

when ,

→ Next year = 41 => 1 * 41 => 2 factors .

→ Present year = 40 => 2³ * 5 => (3 +1)(1+1) => 8 factors .

→ Last year = 39 => 3¹ * 13¹ => (1 + 1)(1 + 1) => 4 factors .

Hence, Present age is equal to 40 years .

Learn more :-

HELP-ASAP-A single line of four cats one behind the other walks towards and head-on into a single line of five cats walk...

https://brainly.in/question/38409325