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Standard Method

From Countdown

The Standard Method is a technique developed by Tony Atkins which finds a solution to three large numbers rounds approximately 60% of the time.

Steps

The method involves working backwards, and as Carol Vorderman used to say: "cover the one you need to use last". That is the one (or possible more than one) number that takes you from the nearest multiple of 25 to the target. Of course it might also be the second nearest multiple of 25 or in extreme cases the third nearest multiple of 25. If the target divides by 25 then there is nothing to cover up. Often it is one or possibly two of the small numbers that you cover, or it may be two of the large ones which produce a small number when they divide into each other.

Then using the remaining small and large numbers get to that multiple of 25, which may involve going via another multiple of 25 then adding or subtracting to get to the required multiple.

Useful things to know are:

  1. The bigs divided: 50 ÷ 25 = 2; 100 ÷ 50 = 2; 75 ÷ 25 = 3; 100 ÷ 25 = 4.
  2. The 25, 50, 75, 100, 125, 150, 175 times tables up to 1050.
    • Quite common are 13 × 75 = 975; 7 × 125 or 5 × 175 = 875; or 8 × 125 = 1000.
    • Also useful is to subtract from the 25 and multiply by 50 (e.g. (25 − 6) × 50 = 950)
  3. Shortcuts like: 25 squared is 625 (this is Tony's favourite sub-method). It can be made from 25 × (100 − 75); 25 × (75 − 50); 25 × 50 ÷ 2; 25 × 75 ÷ 3; 25 × 100 ÷ 4 and more extremely by (50 × 75) ÷ 6 or (50 × 100) ÷ 8.

Some worked examples from Tony's shows:

  • 50 25 75 1 10 7 → 827 — save 50 ÷ 25 = 2; (10 + 1) × 75 = 825; add the 2.
  • 100 25 50 9 7 9 → 577 — save 9 − 7 = 2 (not 100 ÷ 50 etc.); (50 + 25) × 9 = 675; 675 − 100 = 575; add the 2 (He didn't solve this one).
  • 50 25 75 6 4 2 → 623 — save the 2; (75 − 50) × 25 = 625; subtract the 2.
  • 50 100 25 9 4 4 → 252 — save 100 ÷ 50 = 2; 9 + 4 ÷ 4 = 10; 10 × 25 = 250; add the 2.
    • (This can also be done by 4 × 63 or 9 × 28, but he didn't solve this one either).

The two numbers games out of Tony's six that were not soluble with the Standard Method were:

  • 100 75 50 5 6 4 → 878 (only one away possible — 879 by the Standard Method).
  • 50 100 25 7 10 7 → 779 (two possible ways but neither by the Standard Method).

Other methods

Other ways of tackling 3 large when the Standard Method fails include:

  1. Get Up There as Fast as Possible (e.g. (100 + 10) × 7 + 7 + 50 ÷ 25 = 779).
  2. Use the multiplier to make the add on (e.g. (75 + 2 + 1) × 4 + 50 = 362 to make 350 + 12). Using a 1 is often useful to add or subtract before multiplying.
  3. Method of 937.5 (as in 4 large solves) (e.g. (25 × 75 + 1) ÷ 2 = 938 or (50 × 75 + 2) ÷ 4 = 938).
  4. Factors (e.g. (25 − 6) × (75 − 50 − 6) = 361).
  5. Multiply by 10, if you have 10 or 50 ÷ 5 etc., then add/subtract to the target.

Origin

The method was originally posted on an Apterous ticket.