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Q: K divided by 21 divided by N is equivalent to divided by?

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21 N 81 E is India

K Na Li B N

Fertilizers contain especially N, P and K.

K-shell make-up electrons of atom, they have an energy quantum number n = 1.

K, L, M, N, .... shells

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5.25

st37-K+N

The answer depends on if you can choose the same kind of donuts more then once. Or in other words, is repetition permitted. If you can only choose the same kind of donuts only once, it is a 21 choose 12 problem: C(n,k) = n! / (k! (n - k)!) C(21, 12) = 21! / (12! (21 - 12)!) = 21! / (12! (9)!) = 293,930 If you can choose the same kind of donuts more then once, it is a combination with repetition problem. P(n+k-1,k) = (n+k-1)! / (k! (n-1)!) or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12 so C(21+12-1,12) = C(32, 12) = 32! / (12! (32 - 12)!) = 32! / (12! (20)!) = 225,792,840

n = 3Solution:7n/21 = 17n = 1 * 217n = 21n = 21/7n = 3

38 + (n/2) = 21 Therefore, n/2 = -17 n = -17 x 2 n = -34

203 times 21 wtat is the answer

For any natural number k, 21k is a multiple of k.So for example, if k=1, 21 is the first multiple.Now if k=2, we have 42 which is the next one.In fact for any number n, and any natural number k, kn is a multiple of n.

23. If the remainder when a number n is divided by m is r, then it is written as n ≡ r MOD m (read as "n is equivalent to r modulo m") and n = mk + r for some integer k. if n ≡ 643 MOD 837, then n = 837k + 643 (for some k) When n is divided by 31: n ÷ 31 = (837k + 643) ÷ 31 = 837k ÷ 31 + 643 ÷ 31 = 27k + 20 + 23/31 Multiplying both sides by 31: → n = 31 x (27k + 20) + 31 x 23/31 = 31m + 23 (where m = 27k + 20) → n ≡ 23 MOD 31 Thus the remainder when the same number is divided by 31 is 23.

No.K&N is proved to be better than Fram.

Any whole number can be converted to a fraction by putting it over 1.

what is the answer for 42 divided by n?

for (n=1; n<1000; ++n) { for (sum=0, k=1; k<=n/2; ++k) if (n%k==0) sum += k; if (sum==n) printf ("%d\n", n); }

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