{"id":19935,"date":"2013-05-21T21:29:59","date_gmt":"2013-05-21T21:29:59","guid":{"rendered":"http:\/\/thomasjpitts.co.uk\/wp\/?p=19935"},"modified":"2013-05-21T21:29:59","modified_gmt":"2013-05-21T21:29:59","slug":"day-141-an-age-based-maths-solution","status":"publish","type":"post","link":"https:\/\/thomasjpitts.co.uk\/wordpress\/2013\/05\/21\/day-141-an-age-based-maths-solution\/","title":{"rendered":"Day 141: An Age Based Maths Solution"},"content":{"rendered":"

Today, I received this tweet:<\/p>\n

@thomasjpitts<\/a> Explain. RT @uberfacts<\/a>: If you take your age,multiply<\/a> it by 7, then multiply by 1443 the product repeats your age 3 times.<\/p>\n

\u2014 alastair whitelaw (@alastairRP) May 21, 2013<\/a><\/p><\/blockquote>\n

So, what is the maths behind this trick?<\/p>\n

I quite quickly saw that my age produced 303030 (7 x<\/a> 30 = 210, 210 x 1443 = 303030), and thinking backwards, noticed that 7 x 1443 is 10101.<\/p>\n

The trick really only works if you are between 10 and 99 and working backwards works better to explain and turn it into more of a trick.<\/p>\n

For instance, if you asked a person to write their age down three times to create a 6 digit<\/a> number, handed them a calculator and requested they divide it by 1443 then tell you the number they ended up at, assuming you are fairly confident at dividing by 7, you could\u00a0tell\u00a0them their age, thus wowing an audience. Perhaps.<\/p>\n

When you repeat a 2 digit number three times to create a six digit number, you are really multiplying it by 10101.<\/p>\n

Suppose the 2-digit number was 10x<\/em>+y<\/em> (the x<\/em> therefore becomes the number in the tens column and the y<\/em> in the ones column). To multiply by 10101, we can multiply by 10000, 100 and 1, then add them together. The 0s have no effect on the multiplication. This gives:<\/p>\n

10000x<\/em>(10x<\/em>+y<\/em>) = 100000x<\/em> + 10000y<\/em><\/p>\n

100x<\/em>(10x<\/em>+y<\/em>) = 1000x<\/em>\u00a0+ 100y<\/em><\/p>\n

1x<\/em>(10x<\/em>+y<\/em>) = 10x<\/em>\u00a0+ y<\/em><\/p>\n

Adding that lot gives,\u00a0100000x<\/em>\u00a0+ 10000y + <\/em>1000x<\/em>\u00a0+ 100y +\u00a0<\/em>10x<\/em>\u00a0+ y<\/em> or 10101 times the starting number.<\/p>\n

Undoing this using inverse operations requires dividing by 10101. Now, 7 and 1443 are factors of 10101, so asking a volunteer to divide by 1443 does the majority of the work. Some other factors of 10101 are 3, 7, 13 and 37 – so if you can’t divide by 7, you could use other numbers then divide by 3. Somewhat less impressive though.<\/p>\n

Related articles<\/h6>\n