1
2 3 4
5 6 7 8 9
---------------------------
------------------------------
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---------------------------------------till the 10 th row.
Ques 1 . What is the sum of all the numbers in the 10th row of the above pyramid.
Ques 2 . What is the sum of all the numbers on the side of the triangle formed by the 10th row pyramid.
Ques 3. What is the sum of all the numbers inside the triangle formed by the 10th row pyramid.
Ques 4. if ( ab + cd ) ^ 2 = abcd , where ab & cd are two digits number
a) How many values a can take if d = 1 ?
b) How many values b can take if c = 2 ?
c) How many values a can take if b = 1 ?
d) How many such numbers are possible ?
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ReplyDeleteAnswer:-
ReplyDeleteAns for que 1:- 1729
Ans for que 2:- 2226
Ans for que 3:- 2724
ans for que 3 is 2824 i;e 5050-2226= 2824.
ReplyDeleteanswers::
ReplyDeleteans 1= 1729(sum of 82 to 100)
ans 2= 2226
ans 3= 2824(sum of first 100 terms -ans 2)
ans 4(d)= there are 2 such nos. possible i.e.
2025(45*45) ab=20 cd=25
3025(55*55) ab=30 cd=25
also 9801(99*99) satisfies but since 01 is not a two digit no. therefore it is rejected. hence answer is 2.
ans 4(a)= 0
reason :to satisfy above criteria d can't be 1.so a can't take any value.
ans 4(b)= 1
reason: if c=2 then b can take only 1 value i.e. 0. so b can take only 1 value.
ans 4(c) = 0
reason: b can't be 1 as it not satisfied the equation given in the ques. so a can't take any value when b=1.
Q1:-......
ReplyDeleteAnswer:-1729
Q2:-......
Answer:-2226
Q3:-......
Answer:-2824
ans1...
ReplyDeletethe 10th row has no.s 82 through 100.
its sum=sum of 82 to 100
=sum of 1st 100 natrl no.s-sum of 1st 81 natrl no.s
=5050-3321=1729
ans2...
sum of the sides of d triangl individually-the vertices(to avoid doubl countg)
=(295+385+1729)-1-82-100
=2226
ans3...
sum of 1st 100 naturl no.s-ans2
=5050-2226
=2824
ans1......as to how we get 82 the 1st no. of the 10th row...we keep adding successive odd no.s (strtg wid 1) to the numbers dat strt the previous row...eg,the 2nd row strts wid 2(1+1)...,3rd row strts wid 5(2+3)...and so on till we get to the 10th row
ReplyDeleteans1.1729 (sum of 82 to 100)
ReplyDeleteans2.2784 (too long process pls suggest a short cut)
ans3.2266 (sum of 1st 100 terms-ans2)
ans4.(a)0 (no number possible)
ans1=1729
ReplyDeleteans2=2226
ans3=2824
ans4= only 1 value 0f a=9 when d=1.(98+01=99, 99^2=9801)
ans5= only one value of b=0 when c=2. ( 20+25=45, 45^2=2025, 30+25=55, 55^2=3025)
Sol 1 Last number of every row is a perfect square .
ReplyDeleteThus 10th row will end by 100 & 9th row will end by 81 .
thus sum of numbers in 10 th row is
100( 100+1) - 81 ( 81+1)
-------- ---------- = 5050 -3321 = 1729.
2 2
1729 is Ramanujam Number . This is the least positive number which can be expressed as the sum of cubes of two different numbers in two different ways.
1729 = 1728 + 1 = 1000 + 729 = 12^3 + 1^3 = 10^3+9^3
There are infinite such numbers.
Sol 2. 10th row sum 1729
Right side side 1^2 + 2^2 + ---- +9^2 = (1/6)9(9+1)(9X2 + 1)= 285
Left side sum 1^2+1+2^2+1+3^2+1 --- 8^2+1=204+8=212
Required sum 1729+285+212=2226
Sol 3 . Sum of inside numbers Sum of all numbers natural upto 100 - sum of all natural numbers on three sides =
5050 - 2226 = 2824
Sol Numbers possible are
ReplyDelete(20+25)^2 = 2025 & (30+25)^=3025 & (98+01)^2=9801
but 01 is not a two digit number.
Tuus only two possible numbers .
Ramanujan number is 1729.
ReplyDelete1729 is also known as the Hardy – Ramanujan number . This number is also called the Taxicab number.
Ramanujan number is so named after a famous anecdote of the British mathematician G. H. Hardy regarding a hospital visit to the Indian mathematician Srinivasa Ramanujan.
In Hardy's own words:
“I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number...1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two positive cubes in two different ways."
The numbers is such that,
1729 = 1^3 + 12^3 = 9^3 + 10^3
Some observations related to Ramanujan Number–
1. If negative cubes are allowed, 91 is the smallest possible number with similar quality 91 = 6^3 + (−5)^3 = 4^3 + 3^3
2. Interestingly 91 is also a factor of 1729. (91x19=1729)
3. If taking “positive cubes” would not have been a condition, Ramanujan number could have been −91, −189, −1729, and further negative numbers
4. 1729 is also the third Carmichael number and the first absolute Euler pseudoprime. (If you want to know more about this numbers I can discuss it in some other post)
5. Masahiko Fujiwara showed that 1729 is one of four positive integers (with the others being 81, 1458, and the trivial case 1) which, when its digits are added together, produces a sum which, when multiplied by its reversal, yields the original number: 1 + 7 + 2 + 9 = 19; 19 x 91 = 1729
6. Till date only 10 Taxicab numbers are known. Subsequent Taxicab numbers are found using computers.
Sir
ReplyDeletei would like to know about Carmichael and pseudoprime numbers.
Hello Sir,
ReplyDeleteAre you the one who worked with Career Launcher CP, New Delhi?