Math Is Fun Forum

Tazz

Member

Registered: 2007-05-14

Posts: 3

Elementary - but I need a bit of help please.

On some of the workouts this works, but on others not - well, not quite in the same way - can someone please advise on any very simple
' how to ' formulas to get the
same result regardless of the numbers involved ?

Example,
  91 X 92 =
= 9   =8

Subtract the 91 from 100 = 9
                                           X
Subtract the 92 from 100 = 8 = 72 is the last two digits of the result.

Subtract Diagonally,the 9 from 91= 83
     "              "       ,the 8 from 92= 83 so 83 is the first two digits of the result.

Result of  91X 92 = 8372

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Elementary - but I need a bit of help please.

Let the numbers be a = 90+x and b = 90+y

Note that 8000 + 100(x + y) > 100 whereas (100 − a)(100 − b) ≤ 100 (= 100 ⇔ x = y = 0). Hence the last two digits of ab are the product of 100−a and 100−b (or 00 if both numbers are 90).

You can also see from the equation that the first two digits of ab are 80+x+y (or 81 = 80+x+y+1 if x = y = 0, the 1 being the “carry over” from the first part). 80 + x + y = 80 + a − 90 + b − 90 = (a + b) − 100 = b − (100 − a) or a − (100 − b). (Alternatively, you can just add the last digit of both numbers to 80, if this is less confusing than “diagonal” subtraction: for your example, for example, 80 + 1 + 2 = 83; so the first two digits of 91×92 are 83.)

Last edited by JaneFairfax (2007-05-15 22:50:47)

Tazz

Member

Registered: 2007-05-14

Posts: 3

Re: Elementary - but I need a bit of help please.

Yes, thanks very much Jane - The very reason
I included 'Elementary,' and 'Simple,' was to give the impression I was looking for 'E' and 'S,' rather than  X or Y Algebraic equations.

The example I gave can be worked out as a mental ex., and many other such numbers can be multiplied just as easily, but unfortunately not all can be carried out in the same way - I want to use the same basic method as my example, but need to know how to do it regardless of the two numbers.

As you know how to do it algebraically, then that must mean you have the capacity to do it via the simple method - the latter is all I need.

Kind Regards.

Tazz

Member

Registered: 2007-05-14

Posts: 3

Re: Elementary - but I need a bit of help please.

Just looked at   87 X 99 =

In one's mind, subtract the 87 from 100 =13
                     subtract the 99 from 100 = 1
13x1 = 13 = Last two digits of the problem.

Subtract diagonally the 13 from 99 = 86
                             the   1 from 87 = 86
86 is the first two digits of the result.

8613
                      ~~~~~~~~~~~~~~~~~

The problem is, why can't I do the same thing with ?
82 X 88 and others -  what's wrong with maths that it's not consistent ?

What am I missing here please, it's driving me round the bend ?

Last edited by Tazz (2007-05-16 15:26:11)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Elementary - but I need a bit of help please.

No, no. The method only works with numbers between 90 and 99. It won’t for numbers less than 90. If you insist on trying, however, you might need to modify the original method a little.

For 87×99, for instance, x = −3 and y = 9. For the first two digits, subtract from 100 and multiply; you did that and got 13. For the second part, instead of subtracting diagonally, I find it easier to just add x and y to 80; ∴ the first two digits are 80 − 3 + 9 = 86. Thus 83×99 = 8613, what is what you got.

For 82×88, x = −8 and y = −2. Subtracting from 100 and multiplying, 18 × 12 = 216. You keep the 16 (that’s the first two digits) and carry the 2 over to the second part. Second part is* 80 − 8 − 2 + 2 (adding the carried-over 2) = 72. Hence 82×88 = 7216.

*You can also subtract diagonally, but don’t forget to add the 2 that is carried over from the first part:
Subtract 18 from 88 and add 2: 72
Subtract 12 from 82 and add 2: 72

Last edited by JaneFairfax (2007-05-16 22:25:27)