The Last Bean in the Pot

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans. 

A somewhat demented cook removes the beans from the pot, one at a time, according to the following strange rule: 

He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter what color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.
 
At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it?

And the color of last bean is...... 

The Color of Last Bean in the Pot

Little story behind the title!

There are 75 WHITE beans in the pot i.e. they are odd in numbers. Since, they are always taken out in pair, in the end there will be 1 WHITE bean left out.

At some point, when there are 3 bean are left in the pot then there has to be 2 BLACK and 1 WHITE beans in the pot. They can't be 1 BLACK and 2 WHITE beans as for that 73 WHITE beans need to be taken out which is not possible since WHITE are always taken in pair.

So if cook pick 2 BLACK beans (or BLACK & WHITE) at this point then anyhow BLACK and WHITE will be left in the pot. Now, when he pick this pair of BLACK and WHITE then he puts BLACK on the pile and drop WHITE back to the pot as per his rule.

Eventually, WHITE bean will be left in the pot.

MPSC मध्ये विचारला गेलेला प्रश्न

सोडवा गणित

तुम्ही एका यात्रेत आहत,

त्या यात्रेत तुम्हाला काही प्राणी खरेदी करायचे आहेत.त्या प्राण्यांची किमंत खालील प्रमाणे....

*१० रुपयाला - १ हत्ती

* १ रुपयाला - १ घोडा

* १ रुपयाला - ८ उंट

आणि तुमच्याकडे १०० रुपयेच आहेत.प्राण्याची संख्या १०० च आली पाहिजे.

वरील सगळे प्राणी घेणे बंधनकारक आहे.

कसे येतील.???

प्रश्न कठीण वाटतोय का ? उत्तरासाठी येथे क्लिक / टॅप करा.

👉  वाचा -  गोष्ट विषमासूर नावाच्या राक्षसाची !  

MPSC प्रश्नाचे उत्तर : यात्रेतील प्राणी खरेदी

जाणून घ्या काय होता प्रश्न ?

यात्रेमध्ये कोणता प्राणी किती किंमती मध्ये मिळतो ते पाहुयात.

* १० रुपयाला - १ हत्ती

* १ रुपयाला - १ घोडा

* १ रुपयाला - ८ उंट

म्हणजेच १ उंट १/८ रुपयांना मिळतो.

समजा आपण ' क्ष ' हत्ती, ' य ' घोडे आणि ' ज्ञ ' उंट घेतले तर क्ष हत्तींची किंमत होईल १०क्ष, य घोड्यांची किंमत होईल १य आणि ज्ञ उंट (१/८)ज्ञ रुपयांना पडतील. 

आपल्याजवळ १०० रुपये आहेत म्हणजे - 

१०क्ष + य + (१/८)ज्ञ = १००   ......... (१)

आणि आपल्याला एकूण १०० प्राणी घ्यायचे आहेत म्हणजेच -

क्ष + य + ज्ञ = १००

य = १०० - क्ष - ज्ञ          ....... (२)

समीकरण (२) हे (१) मध्ये टाकल्यानंतर,

१०क्ष + (१०० - क्ष - ज्ञ) + (१/८)ज्ञ = १००

९क्ष + १०० - ज्ञ + (१/८)ज्ञ = १००

९क्ष - (७/८)ज्ञ = ०

९क्ष = (७/८)ज्ञ 

क्ष/ज्ञ = ७/७२. 

याचाच अर्थ क्ष = ७ आणि ज्ञ = ७२ व म्हणूनच य = १०० - क्ष - ज्ञ = १०० - ७ - ७२ = २१ असू शकतो.  

म्हणजेच आपण ७० रुपयांचे ७ हत्ती, २१ रुपयांचे २१ घोडे आणि ९ रुपयांचे ७२ उंट खरेदी करायला हवेत जेणेकरून १०० रुपयांमध्ये १०० प्राण्यांची खरेदी पूर्ण होईल.  

एकूण खर्च = ७० + २१ + ९ = १००. 

एकूण प्राणी = ७ + २१ + ७२ = १००.  

One More Alphamatic Problem?

In the following  puzzles, replace the same characters by the same numerals

so that the mathematical operations are correct.
               

Note - Each letter represents a unique digit and vice-versa.

ABCB - DEFC = GAFB

     :          +      -
  DH  x     AB =    IEI
---------------------------
 GGE + DEBB = DHDG

Here is the SOLUTION 

One More Alphamatic Solution!

Look at the problem first!

Rewriting the problem once again,

ABCB - DEFC = GAFB
   :         +       -
  DH x   AB    =    IEI
-------------------------
 GGE + DEBB = DHDG

We have 6 equations from above -

(1) A B C B - D E F C = G A F B  

(2) G G E + D E B B = D H D G  

(3) G A F B - I E I = D H D G 

(4) D E F C +  A B = D E B B 

(5) A B C B : D H  = G G E  

(6) D H x A B = I E I  

Steps :

1. From (1), we have B - C = B. That's possible only when C = 0.

2. If C = 0 then in (1), for tens' place subtraction i.e. C - F = F the carry need to 

    be taken from B. And that subtraction looks like 10 - F = F. Obviously, F = 5.

3. From (3), we see D in result seems to be carry and carry never exceeds 1 

    even if those numbers are 999 + 9999. So, D = 1. 

4. From (1), since C = 0, at hundreds' place (B - 1) - E = A and from (4),

    we have F + A = B (since first 2 digit of first numberremain same in result

    indicating no carry forwarded in addition of FC + AB = BB.

    So placing F = B - A in (B - 1) - E = A gives, F = E + 1. Since, F = 5, then E = 4. 

5. In (3), G at the thousands' place converted to D without actually subtraction 

    of digit from IEI. Since, G and D are different numbers some carry must have been

    taken from G.

    As D = 1 then G = 2.

6. From (1), A - D = G and D = 1 and G = 2 then A = 3 since if carry had been taken 

    from A then A = 4 which is impossible as we already have E = 4. 

7. From (2), E + B = G i.e. 4 + B = 2 only possible with B = 8.

8. With that, in (2), carry forwarded to  G + B = D making it 

    1 + G + B = 1 + 2 + 8 = 11 = 1D  i.e. carry 1 forwarded to G + E = H making it 

    1 + G + E = H = 1 + 2 + 4 = 7.

    Therefore, H = 7 and no carry forwarded as digit D in second number remains

    unchanged in result. 

9. Now (6) looks like - 17 x 38 = 646 = IEI = I4I. Hence, I = 6. 

To sum up,

A = 3, B = 8, C = 0, D = 1, E = 2, F = 5, G = 3, H = 7 and I = 6. 

Eventually, all above 6 equations after replacing digits in place of letters look - 

1. 3808 - 1450 = 2358  ✅

2. 224 + 1488 = 1712   ✅

3. 2358 - 646 = 1712   ✅

4. 1450 + 38 = 1488    ✅

5. 3808 : 17 = 224      ✅

6. 17 x 38 = 646         ✅  

Rewriting in the given format,

3808 - 1450 = 2358
      :         +       -
    17 x     38 =  646
-----------------------------
  224 + 1488 = 1712

Ultimate Test of 3 Logic Masters

Try this. The Grand Master takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the Grand Master's pocket and the two on her own forehead. He asks them in turn if they know the colors of their own stamps:

A: "No."

B: "No."

C: "No."

A: "No."

B: "Yes."

What color stamps does B have? 

'THIS' could be his color combination! 

Earlier logicians had been part of "Spot On The Forehead!" and  "Spot on the Forehead" Sequel Contest