Since 1 is not divisible by 2, 3, 4 or 7 while 1000 is divisible by 5, the answer is 228 1 = 227.
Thanks for the detailed explanation. I needed that!
A three digit number whose digits add up to 9? ... The largest 3 digit number whose digits are all primes is 777. The largest 3 digit number with different prime digits is 753
To maximize the amount of digits, we should make as many digits as we can a 2. That would mean we get 4 2's since we need space for a three. Then we get one three. The largest number we can make with those digits is 32222.
If the sum is less than 220, what is the greatest possible value for the largest of the three integers in the set?: Less than 220, therefore, the number is 212 or the 1st and 3rd digit = 1
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Thanks, you are awesome!
The largest 3 digit palindrome less than 220 is 212. But 212 won't work since it is x+x+1+x+2, so the number has to be divisibe by three. 202 doesn't work either, but 171 does. 3x+3=171, 3x=168 x=56. The largest possibiliyy would be 58.
(-2 + 4i) * i^3 = -2i^3 + 4i^4 = -2 (-1)(i) + 4(1) = 2i + 4 = 4 + 2i ⇒ Note that this lies in Quadrant l
So
-2 + 4i lies in Quadrant ll
Clockwise
Quadrant l
20! = 2^18 * 3^8 * 5^4 * 7^2 * 11 * 13 * 17 * 19
22.5 of the first alloy and 67.5 of the second.
https://www.wyzant.com/resources/answers/192387/a_worker_has_metal_alloy_that_is_30_copper_and_another_hour_if_that_is_55_how_many_kilograms_of_each_alloy_should_the_metal_worker_combine_to_create_6kg_of_45
That's great, you rock!
8.
A prime number will only come up in the factorization of a factorial if the prime is less than n, with n being (n)!. THere are 8 primes less than 20- 2, 3, 5, 7, 11, 13, 17, and 19.
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What about 2100 ???
2100
In every year of the 2000, there is an automatic 2, leaving only a one. There are 3 spots to put it in: 2100, 2010, and 2001. 2100 is the first one after 2010.
Let $M=4a^2 - 2b^2 +a$. Let $j$ be the value of $M$ when $a=5$ and $b=3$, and let $k$ be the value of $M$ when $a=-1$ and $b=4$. Calculate $j+2k$.
j = 4(5)^2 - 2(3)^2 +5 = 100 - 18 + 5 = 87
k = 4(-1)^2 - 2(4)^2 + 5 = 4 - 32 + 5 = -23
j + 2k =
87 + 2(-23) =
87 - 46 =
41
Numbers don't have to be integers, they could be fractional, irrational, ... .
2502 2511 2520 2529 2538 2547 2556 2565 2574 2583 2592 Total = 11 such numbers
yes i believe that is the answer
Nevermind I'm right
A =1.12246 B =1.498307 C =0.435275 D =2 E =1 C < E < A < B < D = C, E, A, B, D
98752 I think
Total number = x
White = (3/8)x
Red = (5/12)x
There are 24 more red than white: Red = White + 24
(5/12)x = (3/8)x + 24
multiply by 24: 10x = 9x + 576
x = 576
