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Consider the following multiplication in decimal notations: (999).(abc)=def132 ,determine the digits a,b,c,d,e,f. solution)a=8b=6c=8d=8e=6f=7In other words, 999 * 877 = 876132.You don''t need to find out d,e,f, as they''re dependent. You only need to find out a,b,c.Now, start with c, because it''s in the ones place and thus easiest. What digit, when multiplied by 9, gives a "2" in the ones? The answer: 8. because 8 * 9 = 72. That the "2" you have there.Now we need to find out b. What digit, when multipled by 9, and added 9 (from the step above, the tens digit), gives a "4" in the ones? The naswer: 6. because 6 * 9 = 54, and added 9 to 4 gives us 13. That''s the "3" we have there. Now we need to find out c. what digit when multiplied with 9 and added 19 (from the step above) , gives a "1" in the ones..? the answer :8 because 8*9 = 72 and added 2 to 19 gives 21. that''s "1" we have therenow we have a=8 , b=6 , c=8 therefore we have 999*868=def132. multiply both the numbers and you get d, e and f..
A computer is programmed to scan the digits of the counting numbers.For example,if it scans 1 2 3 4 5 6 7 8 9 10 11 12 13 then it has scanned 17 digits all together. If the comput
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46+4=
we know that log1 to any base =0 take antilog threfore a 0 =1
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