Teachers, save “Weekly Math Challenge - week of 4/6/20” to assign it to your class.

Teacher Notes (not visible to students)

SLIDE 1: INK = 105, PEN = 630 We have 6 × INK = PEN. This means that I = 1 (if I > 1 then the product would have 4 digits). For the same reason N is at most 6. Also, N has to be even (being the last digit of 6 × K). At the same time, K has to be odd (otherwise K = N, since for an even number X, the numbers X and 6 × X end with the same digit). Finally, K is not equal to 1 (as I = 1). Thus we have three possibilities to consider: N = 0, K = 5; N = 2, K = 7; N = 4, K = 9.They give 6 × 105 = 630, 6 × 127 = 762, 6 × 149 = 894. Only the first one satisfies the condition that different letters stand for different digits. SLIDE 2: The answer can be found by analyzing who wins the game for smaller initial pile sizes (starting from very small and gradually increasing it). The sequence of pile sizes to consider which most quickly leads to the answer is 1,1,1; 0,2,2; 1,2,3; 1,3,4. The first move should be taking 2 matches from the pile of 4, reducing pile sizes to 1,2,3. Then the second player will lose. Whatever move he makes, the first player can do one of three things on his second move (check this!): 1) Win immediately (by taking all matches but 1) 2) Reduce the piles sizes to 1,1,1 3) Reduce the pile sizes to 0,2,2 The positions 1,1,1 and 0,2,2 are clearly winning for the first player. So in situations 2) and 3) he wins on the third move.