CelsiusSolving

Having a units digit of 2 and a remainder of 1 means the units digit of the quotient is 1. So the problem is asking "How many 3 digit integers have a units digit of 1 and are divisible by a number 2-9?"

Because 2, 4, 5, 6, 8 dont have any multiples that end in 1, we can exclude those

Because 9 is a multiple of 3, counting the 3's already includes 9's so we can exclude 9 too

To find the number of 3 digit integers that end in 1 and are divisible by 3 and/or 7, we can see that every 3 terms in the sequence 101, 111, 121, 131, ... is divisible by 3, so that is simply 90 (The number of terms in the sequence)/3 = 30.

To find the number of 3 digit integers that are divisible by 7, we can find that, extending the sequence to 1001, there are 91/7 = 13 terms, and subtracting the 1001 gives 12.

This logic counts integers that are divisible by 21 twice, so subtracting ⌊90/21⌋ = 4 would even it out.

Finally, the sum turns out to be 

30(Multiples of 3) + 12 (Multiples of 7) - 4 (Multiples of 21) = 38 3 digit numbers

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$=

$1-3ab(a-b)$=

$1-3ab$=1

3ab=0,

ab=0

Because ab=0, we know $(a-b)^2$=$a^2+b^2+0$=1

Because of this, $(a+b)^2=1+0=1$, so (a+b) is +/-1

Finally because a+b is 1, and one of either a or b is 0, because ab is 0, the possible pairings of (a,b) are

(0,1), or (1,0)

Hope this helps!

 🤙 🤙 🤙

$9^{x}-2\cdot 3^{x}+1$:

Set a = $3^x$

$a^2-2a+1=0$

$(a-1)^2=0$

a=1, so $3^x$=1

That means x=0 because anything to the 0th power is one

Hope this helps! 

 🤙 🤙 🤙

N mod 10 = a

N mod 2 = b

N mod 20 =?

2 and 10 are not coprime. 10 is multiple of 2.

If a is odd then b=1.

If a is even then b=0.

We don’t need b.

N mod 20 ={a, a+10}