Answer:
I assume you know Arithmetic Progression .
so, we have to find the first and last 4-digit number divisible by 5
first = 1000 , last = 9990
we have a formula,
= a + (n-1)d
here,
is the last 4-digit number divisible by 5.
n is the number of 4-digit even numbers divisible by 5
d is the common difference between the numbers, which is 10 in this case
a is the first 4-digit number divisible by 5
9990 = 1000 + (n-1)*10
899 = n-1
n = 900
Hence, there are 900 4-digit even numbers divisible by 5
I'll do the first 2 and 6, and I challenge you to do the other three on your own!
For 1, from some guess and check we can figure out that 5*5=25. Since 5 is a prime number, that's it!
For 2, we can figure out that 7*7=49 and 7 is a prime number, so we're good there.
From 6, we can do some guess and check to figure out that 2*24=48, 2*12=24, 2*6=12, and 2*3=6, resulting in 2*2*2*2*3=48 since 2 and 3 are prime numbers. We found out, for example, to find 2*12 due to that if 2*24=48, 2*24 is our current factorization. By finding 2*12=24, we can switch it to 2*2*12
Answer:
you need to 12 minus 5b9 as base and don’t forget to mention the integer