A Fistful of Brain Teasers

Replacing the * * * in the previous post, generates the picture above.

The procedure should now be fairly obvious !

Hopefully those little grey cells that were so active 50 years ago are coming back to life ?

Again.......

82 + 2 = 84 and

200 - 164 = 36

.......it's only a matter of routine from here on in !

But i'll do a few more lines just to be sure !

Yep ! we just bring down the next pair of numbers, which again just happen to be "00"

....and we are looking for a single digit to replace those "*"s

Can you figure it out before you sneak a look at the next slide ?

let's try a "4"

844 x 4 = 3376 and that looks promising !

So we'll write it in and then do the usual arithmetic

....and continue ad-infinitum......

....or until we have reached the desired level of accuracy !

.......or you could just revert to Excel.....

and let those old grey cells die off

One for the "quiet hour" at Christmas !

You know what a palindrome number is ? It’s a number that looks the same when read left to right as it does when read right to left. So, 151 is a palindrome but 155 is not a palindrome.

Pick any three-digit, non-palindrome number eg 155 (but you don’t have to pick 155 !!)

Now reverse the number so, 551 in the given example, thus providing a second number.

Subtract the smaller from the larger of these two numbers ie 551 – 155

And write down the answer ie 396 in the example, thus generating a third number.

Now, if this third number has only two digits eg 45, put a zero in front eg 045

Reverse this third number to give a fourth. Again, in the example that would be 693.

Now add the third and fourth numbers together to give 1089…………….

………regardless of the number you started with ! (unless you made a mistake !)

What number am I ?

I’m smaller than 80. I am an odd number. I’m a multiple of 7. And the difference of my two digits is 1.

Now I know this is an incredibly easy teaser. What surprises me, is the incredibly laborious ways in which people tackle it !

So, as well as quoting the answer, could you outline how you tackled the solution ?

21, I simply stepped through the multiples of 7 in my head, got to 21 which met the criteria, then stepped through the rest to confirm no others.

Yes, that's how I did it.

However, i've seen many people start at 80 and work backwards through all the numbers 79, 78, 77, etc highlighting the multiples of seven ie  77, 70, 63....etc. Then continue by eliminating the even numbers that are multiples of seven, then and only then, searching (usually at random) for the one with two digits that are a difference of one.

But as soon as they find 21, they stop, they don't step through any remaining numbers to confirm that 21 is the only solution.

It seems as if the way the question is worded, causes many people to work in a particular way.

I wonder how true this is in other walks of life ?

Lots of pieces of information can cause people to look at a problem and "get confused" I find, rather than looking for a lazy approach to a problem, i.e. the one which will easily help to get to the answer. Many years ago I used to run training courses on fault diagnosis and repair on complex data networks, and the key part of the course was on looking at the system as a whole and working out how to isolate the  faulty areas, not on doing the detailed technical work.

Incidentally the 80 seems to be totally extraneous, the answer's a 2 digit multiple of 7, so the only >80 numbers that meet the criteria are 84, 91 and 98, none of which are odd with a digit difference of 1.

I did it by thinking, more or less instantly, 7 times table up to 77, the odd ones (1,3,5,7,9,11x), needs to digits so not 7, then start looking 21 fits the requirement (35 is difference of 2, no point considering further.

But since about the 1980s (?) learning tables by rote was abandoned in schools, so maybe thta makes a difference - to this and all mental arithmetic.

This programme might be of interest to members of the brainteasers club.

 Alex Bellos heads to Germany to meet the supercalculators taking part in this year's Mental Calculation World Cup. Prepare to be dazzled by some amazing mathematical feats.

 https://www.bbc.co.uk/programmes/m0000qps

Two tungsten filament bulbs of 40W and 60W, each designed for use in a 110V AC power system, are connected in series in an AC power supply of 100V.

Which bulb will glow brighter and why ?

It’s late so might have this wrong... but the 40 Watt bulb will be brighter...

current is constant in a series circuit.

the resistance of the 40 w (@110V) bulb is approx 302 ohms (R = V²/P)

the resistance of the 60 w (@110V) bulb is approx 202 ohms

thrtrfore the current flowing through the 40 and 60 watt bulbs in series (V/R = I) 100/504 = approx 0.2 amps

Now power P = I²R

Therefore in the series circuit Power dissipated in the 40 W bulb is 0.2² x 302 = 12 Watts

Power dissipated in the 60 W bulb is 0.2² x 202 = 8 Watts

Therefore the 40 W bulb will glow brighter....

Hi Simon,

Almost word-for-word the way I tackled this one. Thank you for taking part.

But.............. i'm deeply disappointed that a man of your calibre, even late at night, didn't rigg up a practical experiment to validate the theory.

I'm not sure what effect temperature/heat has on the resistance/luminance of a tungsten filament, but i'm guessing it would be similar in both a 40W and 60W bulb, and thus not change the analysis.

For others browsing, I guess we could mention Kirchhoff's 1st Law as a starting point, together with the various relationships between V, I, R and Power in simple electrical circuits.

Open the Box !

I have three unmarked, identical boxes containing coins.

One box contains two gold coins, the second contains two silver coins and the third, one of each.

Each box is divided and has two lids such that one coin can be seen at a time.

I select one box at random.

I open one lid of said box and reveal a gold coin.

What is the probability that this box is the one with the gold and silver coins

1/3

1/2 (it cannot be the box with 2 silver coins)

There are three permutations that give you a gold coin, but only one of them gives you the silver coin too.

One side of the box contains a gold coin. So the silver/silver is out of contention. The second lid will be gold if gold/gold and silver if silver/gold. There is no other possibility, so simply a 50:50 chance at that point.

Maybe I’m being too simplistic?

The probability of the second coin being also gold is 2/3, therefore it being silver must be 1/3. It’s the opposite of the standard solution to the Bertrand’s box paradox, which is rather like that ‘do I switch to the other cup’ Monty Hall problem. My annoyingly good at maths son tries to explain it to us, invariably after we have had too much wine. 

Innocent Bystander posted:

One side of the box contains a gold coin. So the silver/silver is out of contention. The second lid will be gold if gold/gold and silver if silver/gold. There is no other possibility, so simply a 50:50 chance at that point.

Maybe I’m being too simplistic?

+One.

Can't be anything else.