Help/Hint required.

Q1)Find the number of 4 digit numbers with distinct digits chosen from the set {0, 1, 2, 3, 4, 5} in which no two adjacent digits are even.

I'm not sure about the answer. I'm getting 60. There's a high chance I may be wrong.

Q2) Let a, b, c > 0. If 1c,1band1a\dfrac 1 c , \dfrac 1 b \quad and \dfrac 1 a are in arithmetic progression, and if a2+b2,b2+c2,c2+a2a^2+b^2,b^2+c^2,c^2+a^2 are in a geometric progression, prove that a=b=c

Somehow, I managed to get a=b=c. but I'm not really confident of my approach.

Q3) Let n be a positive integer such that 2n + 1 and 3n + 1 are both perfect squares. Show that 5n + 3 is a composite number.

I tried this

Let 2n+1=k22n+1=k^2 and 3n+1=m23n+1=m^2

5n+3=m2+k2+15n+3=m^2+k^2+1

Then I used modulo to bash it out. am I on the right track? Because this doesn't seem to lead to the answer.

Thanks!

Note by Mehul Arora
5 years, 9 months ago

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1 vote

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Comments

Q1) Let the odd digit be OO and even digit be EE.So by product rule we can work out each case. So hereby I give the possible combinations:

OEOO=3×3×2×1=18OOEO=3×2×3×1=18EOOO=2×3×2×1=12OOOE=3×2×1×3=18EOOE=2×3×2×2=24OEOE=3×3×2×2=36EOEO=2×3×2×2=24⇒18+18+12+18+24+36+24=150OEOO = 3\times 3 \times 2 \times 1 = 18 \\ OOEO = 3\times 2 \times 3 \times 1 =18 \\ EOOO = 2\times 3 \times 2 \times 1 = 12 \\ OOOE = 3\times 2 \times 1 \times 3 = 18 \\ EOOE = 2\times 3 \times 2 \times 2 = 24 \\ OEOE = 3\times 3 \times 2 \times 2 = 36 \\ EOEO = 2\times 3 \times 2 \times 2 = 24 \\ \Rightarrow 18+18+12+18+24+36+24 = \boxed{150}

Nihar Mahajan - 5 years, 9 months ago

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Thanks Nihar! :D

Mehul Arora - 5 years, 9 months ago

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Saale dost ko thanks bolta hain :P

Nihar Mahajan - 5 years, 9 months ago

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@Nihar Mahajan – Ahahahahahh xD

Chal nahi bolta thanks xD

Waapas liya thanks xD

Mehul Arora - 5 years, 9 months ago

For the record, here are the numbers:

1
3210,,5210,,1230,,5230,,1250,,3250,,3410,,5410,,1430,,5430,,1450,,3450,,3012,,5012,,3412,,5412,,1032,,5032,,1432,,5432,,1052,,1452,,3052,,3452,,3014,,5014,,3214,,5214,,1034,,5034,,1234,,5234,,1054,,3054,,1254,,3254,,5031,,5231,,5431,,3051,,3251,,3451,,5013,,5213,,5413,,1053,,1253,,1453,,3015,,3215,,3415,,1035,,1235,,1435,,2301,,2501,,2341,,2541,,2103,,2503,,2143,,2543,,2105,,2145,,2305,,2345,,4301,,4501,,4321,,4521,,4103,,4503,,4123,,4523,,4105,,4305,,4125,,4325,,5301,,3501,,5321,,3521,,5341,,3541,,5103,,1503,,5123,,1523,,5143,,1543,,3105,,3125,,3145,,1305,,1325,,1345,,2310,,2510,,2130,,2530,,2150,,2350,,2314,,2514,,2134,,2534,,2154,,2354,,4310,,4510,,4130,,4530,,4150,,4350,,4312,,4512,,4132,,4532,,4152,,4352,,2531,,2351,,2513,,2153,,2315,,2135,,4531,,4351,,4513,,4153,,4315,,4135,,5310,,3510,,5130,,1530,,3150,,1350,,5312,,3512,,5132,,1532,,3152,,1352,,5314,,3514,,5134,,1534,,3154,,1354,

Agnishom Chattopadhyay - 5 years, 9 months ago

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Comp. Science Gawd is here. _/_

Mehul Arora - 5 years, 9 months ago

For the 2nd one, We get that b=2ac/a+cb=2ac/a+c Substitute this in the second equation, (b2+c2)/(a2+b2)=(c2+a2)/(b2+c2)( {b}^2+{c}^2)/({a}^2+{b}^2)=({ c}^2+{a}^2)/({b}^2+{c}^2) By simplifying we get, a=b=c

Saarthak Marathe - 5 years, 9 months ago

Wait in Q1 when there asking for no two adjacent digits are even, doesn't that mean none of the digits are even ? O.O

Alan Yan - 5 years, 9 months ago

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Not necessarily I guess. The placement of the digits could be OEOO

Mehul Arora - 5 years, 9 months ago

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Oh, so no two adjacent digits are both even?

Alan Yan - 5 years, 9 months ago

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@Alan Yan – Exactly :P

Mehul Arora - 5 years, 9 months ago

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@Mehul Arora – In that case I got 150 as an answer :(.

550 doesn't make sense because the maximum number of four digit numbers with distinct digits from the given set is 5⋅5⋅4⋅3=3005 \cdot 5 \cdot 4 \cdot 3 = 300 , which is less than your answer.

Alan Yan - 5 years, 9 months ago

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@Alan Yan – Oh! I guess i mistook the question for a,b,c and d not necessarily being distinct. Lemme work this one out. Till then, please work the other questions :)

Mehul Arora - 5 years, 9 months ago

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@Mehul Arora – Well I just took it in cases. You need at least one even and at most two evens.

So you just work from there.

Alan Yan - 5 years, 9 months ago

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@Alan Yan – Oh okay, thanks! :D

Mehul Arora - 5 years, 9 months ago

@Alan Yan – I got 60 as the answer. Could you please elaborate on how you got 150?

Mehul Arora - 5 years, 9 months ago

Oh, I like question 3. The main way to show that a number is composite, is to factorize it and then show that the factors are not ±1\pm1

Calvin Lin Staff - 5 years, 9 months ago

1) 278

Avanthi Vasudevan - 5 years, 9 months ago
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