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Question Number 176155 by adhigenz last updated on 14/Sep/22

$$\mathrm{Find}\mathrm{how}\mathrm{many}\mathrm{distinct}\mathrm{integers}\mathrm{are}\mathrm{there}\mathrm{in}\mathrm{this}\mathrm{sequence}:$$$$\lfloor \frac{1^2+1}{100}\rfloor ,\lfloor \frac{2^2+2}{100}\rfloor ,\lfloor \frac{3^2+3}{100}\rfloor ,...,\lfloor \frac{{100}^2+100}{100}\rfloor$$$$\mathrm{where}\lfloor x\rfloor \mathrm{is}\mathrm{the}\mathrm{greatest}\mathrm{integer}\mathrm{that}\mathrm{is}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}x$$

Commented by Rasheed.Sindhi last updated on 14/Sep/22

$$102?$$

Commented by mr W last updated on 14/Sep/22

$$therearetotally100integers.sothe$$$$numberofdistinctintegersmustbe$$$$lessthan100.$$

Commented by Rasheed.Sindhi last updated on 14/Sep/22

$${You}^{\prime }reveryrightsir!I^{\prime }vecommitted$$$$mistake.$$

Answered by Rasheed.Sindhi last updated on 14/Sep/22

$$\lfloor \frac{1^2+1}{100}\rfloor ,\lfloor \frac{2^2+2}{100}\rfloor ,\lfloor \frac{3^2+3}{100}\rfloor ,...,\lfloor \frac{{100}^2+100}{100}\rfloor$$$$$$$$\lfloor \frac{n^2+n}{100}\rfloor =\lfloor \frac{\frac{n(n+1)}{2}}{\frac{100}{2}}\rfloor =\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor$$$$$$$$0⩽\frac{n(n+1)}{2}<50;n=1,2,...,9:$$$$\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor =0$$$$50⩽\frac{n(n+1)}{2}<100;n=10,11,...,13:$$$$\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor =1$$$$....$$$$...$$$$50k⩽\frac{n(n+1)}{2}<50(k+1):$$$$\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor =k$$$${We}^{\prime }vetosolvemanysuch$$$$inequalities...$$$$Continue...$$

Answered by Rasheed.Sindhi last updated on 16/Sep/22

$$\ll \underset{◼\mathbf{A}\mathbf{N}\mathbf{S}\mathbf{W}\mathbf{E}\mathbf{R}◼}{‘‘\mathbf{SOMETHING}\mathbf{is}\mathbf{better}\mathbf{than}{\mathbf{NOTHING}}^{″}}\gg$$$$\lfloor \frac{1^2+1}{100}\rfloor ,\lfloor \frac{2^2+2}{100}\rfloor ,\lfloor \frac{3^2+3}{100}\rfloor ,...,\lfloor \frac{{100}^2+100}{100}\rfloor$$$$$$$$\lfloor \frac{n^2+n}{100}\rfloor =\lfloor \frac{\frac{n(n+1)}{2}}{\frac{100}{2}}\rfloor =\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor$$$$$$$$50k⩽\frac{n(n+1)}{2}<50(k+1),k\in \{0,1,2,...\}:$$$$\lfloor \frac{\frac{n(n+1)}{2}}{50}\rfloor =k$$$$For\frac{N}{D\in \mathbb{N}}tobeinteger,Nshouldbe$$$$integralmultipleofD$$$$\begin{array}{|cc|}\hline n& \underset{\stackrel{\mathrm{OR}}{cumulativetotal}}{n(n+1)/2}\\ 1& 1\\ 2& 3\\ 3& 6\\ 4& 10\\ 5& 15\\ 6& 21\\ 7& 28\\ 8& 36\\ 9& 45<50\left(1\right)\\ 10& 55\\ 11& 66\\ 12& 78\\ 13& 91<100\left(2\right)\\ 14& 105\\ 15& 120\\ 16& 136<150\left(3\right)\\ 17& 153\\ 18& 171\\ 19& 190<200\left(4\right)\\ 20& 210\\ 21& 231<250\left(5\right)\\ 22& 253\\ 23& 276<300\left(6\right)\\ 24& 300\\ 25& 325<350\left(7\right)\\ 26& 351\\ 27& 378<400\left(8\right)\\ 28& 406\\ 29& 435<450\left(9\right)\\ 30& 465\\ 31& 496<500\left(10\right)\\ 32& 528<550\left(11\right)\\ 33& 561\\ 34& 595<600\left(12\right)\\ 35& 630<650\left(13\right)\\ 36& 666<700\left(14\right)\\ 37& 703\\ 38& 741<750\left(15\right)\\ 39& 780<800(16\\ 40& 820<850\left(17\right)\\ 41& 861<900\left(18\right)\\ 42& 903\\ 43& 946<950\left(19\right)\\ 44& 990<1000\left(20\right)\\ 45& 1035<1050\left(21\right)\\ 46& 1081<1100\left(22\right)\\ 47& 1128<1150\left(23\right)\\ 48& 1176<1200\left(24\right)\\ 49& 1225<1250\left(25\right)\\ 50& 1275<1300\left(26\right)\\ \hline\end{array}\begin{array}{|cc|}\hline n& \underset{\stackrel{\mathrm{OR}}{cumulativetotal}}{n(n+1)/2}\\ 51& 1326<1350\left(27\right)\\ 52& 1378<1400\left(28\right)\\ 53& 1431<1450\left(29\right)\\ 54& 1485<1500(30\\ 55& 1540<1550\left(31\right)\\ 56& 1596<1600\left(32\right)\\ 57& 1653<1700\left(33\right)\\ 58& 1711<1750\left(34\right)\\ 59& 1770<1800\left(35\right)\\ 60& 1830<1850\left(36\right)\\ 61& 1891<1900\left(37\right)\\ 62& 1953<2000\left(38\right)\\ 63& 2016<2050\left(39\right)\\ 64& 2080<2100\left(40\right)\\ 65& 2145<2150\left(41\right)\\ 66& 2211<2250\left(42\right)\\ 67& 2278<2300\left(43\right)\\ 68& 2346<2350\left(44\right)\\ 69& 2415<2450\left(45\right)\\ 70& 2485<2500\left(46\right)\\ 71& 2556<2600\left(47\right)\\ 72& 2628<2650\left(48\right)\\ 73& 2701<2750\left(49\right)\\ 74& 2775<2800\left(50\right)\\ 75& 2850<2900\left(51\right)\\ 76& 2926<2950\left(52\right)\\ 77& 3003<3050\left(53\right)\\ 78& 3081<3100\left(54\right)\\ 79& 3160<3200\left(55\right)\\ 80& 3240<3250\left(56\right)\\ 81& 3321<3350\left(57\right)\\ 82& 3403<3450\left(58\right)\\ 83& 3486<3500\left(59\right)\\ 84& 3570<3600\left(60\right)\\ 85& 3655<3700\left(61\right)\\ 86& 3741<3750\left(62\right)\\ 87& 3828<3850\left(63\right)\\ 88& 3916<3950\left(64\right)\\ 89& 4005<4050\left(65\right)\\ 90& 4095<4100\left(66\right)\\ 91& 4186<4200\left(67\right)\\ 92& 4278<4300\left(68\right)\\ 93& 4371<4400\left(69\right)\\ 94& 4465<4500\left(70\right)\\ 95& 4560<4600\left(71\right)\\ 96& 4656<4700\left(72\right)\\ 97& 4753<4800\left(73\right)\\ 98& 4851<4900\left(74\right)\\ 99& 4950<5000\left(75\right)\\ 100& \underset{\overline{)\begin{array}{c}76\end{array}}}{5050<5100\left(76\right)}\\ \hline\end{array}$$$$Changingcolouroffonthere$$$$means{}^{\prime }otherdistinct{integer}^{\prime }.$$

Commented by mr W last updated on 15/Sep/22

$$\mathrm{sum}(\mathrm{if}(\mathrm{ceil}\left(\mathrm{a}\left(\mathrm{k}\right)\right)<\mathrm{a}(\mathrm{k}+1)),\mathrm{k}=0..101)$$

Commented by Rasheed.Sindhi last updated on 15/Sep/22

$$Ans:76$$$$adhigenzsir,pleaseconfirmthe$$$$answerbeforeIgivefinalshapeto$$$$mysolution.$$

Commented by mr W last updated on 15/Sep/22

$$answer76iscorrectsir.$$

Commented by Rasheed.Sindhi last updated on 15/Sep/22

$$\mathbb{T}\mathbf{h}\mathbf{a}\mathbf{n}\mathbb{k}\mathbf{s}\mathbf{sir}!$$

Commented by mr W last updated on 15/Sep/22

$$\lfloor \frac{n^2+n}{100}\rfloor =k$$$$1⩽n⩽100$$$$\Rightarrow 0⩽k⩽101$$$$butnoteachnumberfrom0to101is$$$$forkpossible.$$$$$$$$k⩽\frac{n(n+1)}{100}<k+1$$$$100k⩽n(n+1)<100(k+1)$$$$n^2+n-100k⩾0$$$$n⩾\frac{-1+\sqrt{1+400k}}{2}$$$$n^2+n-100(k+1)<0$$$$n<\frac{-1+\sqrt{1+400(k+1)}}{2}$$$$\frac{-1+\sqrt{1+400k}}{2}⩽n<\frac{-1+\sqrt{1+400(k+1)}}{2}$$$$wecanseea‘‘k^{″}ispossible,i.e.a‘‘n^{″}$$$$exists,onlyif$$$$\lceil \frac{-1+\sqrt{1+400k}}{2}\rceil <\frac{-1+\sqrt{1+400(k+1)}}{2}$$$$forkfrom0to101,only76numbers$$$$fulfillthiscondition.icheckedthis$$$$usingGraph.$$

Commented by mr W last updated on 15/Sep/22

Commented by mr W last updated on 15/Sep/22

$$youdidaniceworksirusingthetable!$$

Commented by Rasheed.Sindhi last updated on 15/Sep/22

$${\mathbb{T}}_{}^{{\mathbb{H}}^{\mathbb{A}}\mathbb{N}}\mathbb{X}$$$${\mathcal{S}}_{\mathcal{I}}\mathcal{R}$$

Commented by Tawa11 last updated on 18/Sep/22

$$\mathrm{Great}\mathrm{sirs}$$

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