(A) Use two digits to make the smallest possible positive integer. (B) Five 3s can express 37: 37=33+3+3/3 Find another way to do it. (C) Use six identical digits to make 100. (Several solutions are possible.) (D) Use five 4s to make 55. (E) Use four 9s to make 20.

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I wear you down, yet you’ll mourn me once I fly. You can kill me, but I will never die. What am I?

Solve A+B+B+B+80+90=100 with No negative numbers

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(A) Use two digits to make the smallest possible positive integer. (B) Five 3s can express 37: 37=33+3+3/3 Find another way to do it. (C) Use six identical digits to make 100. (Several solutions are possible.) (D) Use five 4s to make 55. (E) Use four 9s to make 20.

Answer: (A) 1 X 1; 1/1;2/2;ect....;1-0;2-1;and many others. (B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3 (C) 99 + 99/99; 55+55- 5- 5; (666-66)/6 (D) 44 + 44/4=55. (E) 9 + 99/9=20.

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(A) 1 X 1; 1/1;2/2;ect….;1-0;2-1;and many others.
(B) 37=333/3X3; 37=3 X 3 X 3 + 3/.3
(C) 99 + 99/99; 55+55- 5- 5; (666-66)/6
(D) 44 + 44/4=55.
(E) 9 + 99/9=20.

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Solve A+B+B+B+80+90=100 with No negative numbers

Answer: A is log[1/(10^10)] and B is log[1/(10^20)]

Solution:

Mathematical logical Solution:

log [1/(10^10)] = -10 log(10) = -10 ->A

log [1/(10^20)] = -20 log(10) = -20 ->B

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So, LHS: A + 3B + 80 + 90

=> -10 + 3(-20) + 80 + 90

=> 100 -> RHS

Hence proved! ( No negative numbers though)

Hence, A is log[1/(10^10)] and B is log[1/(10^20)]

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