How many integers between 1 and 21,000, including both 1 and 21,000, are divisible by 2, 3 or 7 but not 5?

Answer 1

ALL: 21000 = A+B+C+D+E+F+G+H+I+J+K+L+M+N+O+P

by 2: 10500 = A+B+D+G+H+J+L+N

  NOT by 2: 21000 – 10500 = 10500 = C+E+F+I+K+M+O+P

by 3: 7000 = B+C+E+G+H+J+K+M

  NOT by 3: 21000 – 7000 = 3000 = A+D+F+I+L+N+O+P

by 5: 4200 = D+F+G+J+K+M+N+O

  NOT by 5:  21000 – 4200 = 16800 = A+B+C+E+H+I+L+P

by 7: 3000 = E+H+I+J+L+M+N+O

   NOT by 7: 2100 – 3000 = A+B+C+D+F+G+K+P

by 2 and 3 (that is, 6):  2800 = B+G+H+J

   NOT by 6: 21000 – 2800 = 18200 = A+C+D+E+F+I+K+L+M+N+O+P

by 2 and 5 (that is, 10): 2100 = D+G+J+N

   NOT by 10: 21000 – 2100 = 18900 = A+B+C+E+F+H+I+K+L+M+O+P

by 2 and 7 (that is, 14): 1500 = H+J+L+N

   NOT by 14: 21000 – 19500 = A+B+C+D+E+F+G+I+K+M+O+P

by 3 and 5 (that is, 15): 1400 = G+J+K+M

   NOT by 15: 21000 – 1400 = A+B+C+D+E+F+H+I+L+N+O+P

by 3 and 7 (that is, 21): 1000 = E+H+J+M

   NOT by 21: 21000 – 1000 = A+B+C+D+F+G+I+K+L+N+O+P

by 5 and 7 (that is, 35): 600 = J+M+N+O

   NOT by 35: 21000 – 600 = 20400 = A+B+C+D+E+F+G+H+I+K+L+P

by 2 and 3 and 5 (that is, 30): 700 = G+J

   NOT by 30: 21000 – 700 = A+B+C+D+E+F+H+I+K+L+M+N+O+P

by 2 and 3 and 7 (that is, 42): 500 = H+J

   NOT by 42: 21000-500 = 20500 = A+B+C+D+E+F+G+I+K+L+M+N+O+P

by 2 and 3 and 5 and 7 (that is, 210): 100 = J

   NOT by 210: 21000 – 100 = 20900 = A+B+C+D+E+F+G+H+I+K+L+M+N+O+P

The problem asks for: "divisible (by 2 or 3 or 7) but[and] not 5?"

   This is: A+B+C+E+H+I+L

J=100

G=600

H=400

    2800 = B+G+H+J

B=2400

   1400 = (G+J)+(K+M) = 700 + K + M; so, (K+M)=700

    7000 = B+C+E+G+H+J+K+M = 2400+(C+E)+600+400+100+(700); so (C+E)=2400.

. . .

A=4800, B=2400, C=2400, D=1200, E=400, F=1200, G=600, H=400, I=800, J=100, K=600, L=800, M=100, N=200, P=4800

A+B+C+E+H+I+L = 12000

Answer 2

2, 3, 7, and 5 are all prime and factors of 21000.

Thus, the solution is 1/2 * 1/3 * 1/7 * (1 - 1/5) * 21000 = 4/210 * 21000 = 400