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Alok is attending a workshop "How to do more with less" and today's theme is “Working with fewer digits”. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fewer digits. The problem posed at the end of the workshop is “How many 7 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?” Can you help Alok find the answer?
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- last 2 digits we can hv only 12, 24,32,44 or 52 (since divisible by 4),
5 place 5 digits, way of arrange 5^5
ans will be 5*5^5=5^6 - 12 years agoHelpfull: Yes(30) No(0)
- ans is 5^6
- 12 years agoHelpfull: Yes(29) No(4)
- 4*5^5
5^5 for first 5 digits and at the last 2 digits we can hv only 12, 24,32 or 52 (since divisible by 4), der r 4 ways 2 arrange, so ans is 4*5^5. - 12 years agoHelpfull: Yes(3) No(7)
- ans : 5^6.
- 12 years agoHelpfull: Yes(2) No(1)
- ans is 5^6
- 12 years agoHelpfull: Yes(2) No(1)
- sory..... it was 5^6 coz 44 should also be considered.
- 12 years agoHelpfull: Yes(1) No(1)
- here we have to develop those no. which are dividable by 4. For this last two digit no must be dividable by 4. That can only be in this case is (12,24,32,44,52)
now we have to design the starting five digit that can be taken as
5p5 that is 5*4*3*2*1 = 120
now total no. that can be formed is 120*5 = 600.
so Ans is 600.
Thanks - 10 years agoHelpfull: Yes(1) No(0)
- please read the question correctly
wrong question!
7 digit number can not be formed using 5 digits only. - 12 years agoHelpfull: Yes(0) No(17)
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