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
3 years, 11 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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=2418+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 - 3 years, 11 months ago

Log in to reply

Thanks Nihar! :D

Mehul Arora - 3 years, 11 months ago

Log in to reply

Saale dost ko thanks bolta hain :P

Nihar Mahajan - 3 years, 11 months ago

Log in to reply

@Nihar Mahajan Ahahahahahh xD

Chal nahi bolta thanks xD

Waapas liya thanks xD

Mehul Arora - 3 years, 11 months ago

Log in to reply

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 Staff - 3 years, 11 months ago

Log in to reply

Comp. Science Gawd is here. _/_

Mehul Arora - 3 years, 11 months ago

Log in to reply

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 - 3 years, 11 months ago

Log in to reply

Log in to reply

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 - 3 years, 11 months ago

Log in to reply

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

Mehul Arora - 3 years, 11 months ago

Log in to reply

Oh, so no two adjacent digits are both even?

Alan Yan - 3 years, 11 months ago

Log in to reply

@Alan Yan Exactly :P

Mehul Arora - 3 years, 11 months ago

Log in to reply

@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 5543=3005 \cdot 5 \cdot 4 \cdot 3 = 300 , which is less than your answer.

Alan Yan - 3 years, 11 months ago

Log in to reply

@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 - 3 years, 11 months ago

Log in to reply

@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 - 3 years, 11 months ago

Log in to reply

@Alan Yan Oh okay, thanks! :D

Mehul Arora - 3 years, 11 months ago

Log in to reply

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

Mehul Arora - 3 years, 11 months ago

Log in to reply

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 - 3 years, 11 months ago

Log in to reply

1) 278

Avanthi Vasudevan - 3 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...