How Good Is Your Maths?

[tamborine]

http://www.themathsfactor.com/mathsmadeeasy/?awc=5757_1416503077_d53a1aff2109da443f9a5a2c295f84a1&utm_source=Affiliate&;utm_medium=Referral&utm_campaign=PRI_MAT_27JAN2014_TMF_General&source=aw

[B00]

well i'll be honest along with you Tambo ;-)

I only got 5.

[-Talbot-]

9........... I narrowed the population one down to B or C and got it wrong.
I also narrowed down q10 and got it right.

My 11 year old son got 9 but he did far quicker than me. (he was one off an early bedtime)

[Cloverjo]

BlackadderV. CV appeared on a tv programme recently (Would I lie to You, I think) on which she said - truthfully, apparently - that she often worked out the maths puzzle so quickly she had time to play jokes on the production staff while the clock was counting down. I think I believe you, though.

Incidentally, I wonder if Rachel Riley uses this sometimes.
http://incoherency.co.uk/countdown/
She often says that a game is impossible to solve, but how can she prove it in the time allowed - or at all - without a computer program that can go through every permutation of numbers and operation?

[andy-hughes]

I got 7, which considering I failed GCE maths three times at Grade 9, and that was forty-four years ago, I am quite happy with that!

[jim360]

It's all right.

[BlackadderV]

Clover, yes. Sometimes she may have done, but at others she would be struggling. I refuse to believe that anyone could be so quick 100% of the time.

[-Talbot-]

If she did cheat why did Carol V sometimes say "I think it's possible you will have to come back to me"?

Then do it when RW got back to her on the letters round.

[jim360]

I think Rachel Riley's mental maths is reputed to be excellent. In which case she probably knows all the tricks needed to see whether or not a target is possible without having to do much calculation. This isn't that unreasonable. For example, given the numbers 2, 5 and 10, usable only once each at most, it's clear that there is no way to get above 100 (2*5*10), and most numbers above 50 would be impossible too (apart from 52 and 70). With experience, you could probably perform similar tricks for six numbers and a target of no greater than 1000.

Alternatively, it's possible that she might have gone backwards to some extent. Imagine being given a target of 773. How many ways are there to make this with six numbers? It might well be that you can prove that there's no way to do it unless you have a 7, or some way to make a 7, so a quick inspection of the numbers 2, 4, 8, 10, 25, 100 will show that (if it were true that you needed a 7) the target is impossible. Again, that is the sort of thing that might come with practice.

I have made that last example up, so if anyone can find 773 given those numbers I'd not be surprised.

In the end I suppose it just comes down to experience.

[Cloverjo]

Ok, thanks Jim. However, sometimes she says the best answer is 1 away from the target, which also happens to be the best answer that the calculator link I posted earlier gives.
I take your point though.

[jennyjoan]

I got 7

[jno]

Dizmo, that one didn't need paper and pen as the possible answers were well separated.

You just have to estimate: 72 ÷ 8 is 9. 45 – 37 is 8. Add 8 and 9 and you get 17. It would have been much trickier if all the numbers were close to 17, but only one was very close.

[Tilly2]

9/10. Umph!

[Maydup]

I scored 10/10 but it was touch and go without a pen and paper. Phew.

[BlackadderV]

I think a knowledge of prime factors might help. As 773 seems to be a prime number, though I have not tested it, a solution with Jim's given numbers looks nigh impossible.

[jim360]

It's actually not hard to get to, which illustrates how phenomenally good you have to be to see when it's not possible to solve a numbers game.

8*100 = 800
25+2 = 27 (there, of a sort, is the 7 I mentioned you might need!)
800 - 27 = 773.

[BlackadderV]

Dohhhhh! I was looking for the hard way, not the easy way.

[Slapshot]

10

[Caran]

8/10, pleased with that.