--- 1 ---
991 * 991 = ?
991 = 1000 - 9; keep 9 in your mind
991 = 1000 - 9; keep 9 in your mind
991 * 991 = (991-9) & ...(9*9) = 982081
…
991 * 999 = ?
991 = 1000 - 9; keep 9 in your mind
999 = 1000 - 1; keep 1 in your mind
991 * 999 = (991-1) & ...(1*9) = 990009
--- 2 ---
1001 * 1001 = ?
1001 = 1000 + 1; keep 1 in your mind
1001 = 1000 + 1; keep 1 in your mind
1001 * 1001 = (1001 + 1) & ...(1*1) = 1002001
…
1001 * 1009 = ?
1001 = 1000 + 1; keep 1 in your mind
1009 = 1000 + 9; keep 9 in your mind
1001 * 1009 = (1001 + 9) & ...(1*9) = 1010009
In those post, we employed 1000 as our ref number. As we evaluated 991 as a number which is 9 units less than 1000, or for 1001, we said it is one unit more than 1000. So in the last two posts, 1000 had an important role for our calculations; hence, we call it as a ref number.
In “Race a Calculator! [Part One]”, we learned to multiply, in head, two numbers which both were less than the ref number and also very close to it. In “Race a Calculator! [Part Two]”, we learned to multiply two numbers which both were more than the ref number and also very close to it. In this post, we examine multiplying two numbers which are close to a ref number; however, one more and one less than the ref number.
Consider the following multiplications and try to find a pattern:
991 * 1001 = 991991
992 * 1001 = 992992
993 * 1001 = 993993
...
991 * 1002 = 992982
992 * 1002 = 993984
993 * 1002 = 994986
...
999 * 1009 = 1007991
A pattern can be inferred; it is somehow a combination of what we found in the last two posts.
For 999 * 1009:
999 = 1000 – 1 (keep -1 in your mind)
1009 = 1000 + 9 (keep +9 in your mind)
then:
999 * 1009
= (999 + 9) & … (-1 * 9)
= (1008) & 00[-9]
= 1007 & 100[-9] (take 1 from left side and give a thousand to the right side because we cannot add -9 to “000” to get a positive number)
= 1007 & 991
= 1007991
Let’s try another multiplication; e.g. 994 * 1007:
994 * 1007
994 = 1000 – 6 (keep -6 in your mind)
1007 = 1000 + 7 (keep +7 in your mind)
994 * 1007
= (994 + 7) & … (-6 * 7)
= (1001) & 0[-42]
= 1000 & 10[-42] (take 1 from left side and give a thousand to the right side because we cannot add -42 to “000” to get a positive number)
= 1000 & 958
= 1000958
You can apply this pattern for larger or smaller numbers (follow the next posts for all the ranges of numbers); for example, let's apply it for 999994 * 1000007 (as we did for 994 * 1007).
For 999994 * 1000007, we select 1000000 as the ref number; so:
999994 = 1000000 - 6; keep -6 in your mind
1000007 = 1000000 + 7; keep +7 in your mind
999994 * 1000007
= (999994 + 7) & … (-6 * 7)
= (1000001) & 0000[-42]
= 1000000 & 1000[-42] (take 1 from left side and give a million to the right side because we cannot add -42 to “000000” to get a positive number)
= 1000000 & 999958
= 1000000999958
Do the following exercises without a pen and paper, or any calculator; just in the head:
1) 993 * 995 (review Race a Calculator! [Part One] if you cannot remember the trick)
2) 1008 * 1006 (review Race a Calculator! [Part Two] if you cannot remember the trick)
3) 997 * 1003
4) 10005 * 9999
5) 9995 * 10008
6) 10100 * 9997
7) 9999993 * 10000008
8) 999999992 * 1000000004
[continued ...]
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