Find-how-many-distinct-integers-are-there-in-this-sequence-1-2-1-100-2-2-2-100-3-2-3-100-100-2-100-100-where-x-is-the-greatest-integer-that-is-less-than-

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

Find how many distinct integers are there in this sequence :

⌊ \frac{1^{2} + 1}{100} ⌋, ⌊ \frac{2^{2} + 2}{100} ⌋, ⌊ \frac{3^{2} + 3}{100} ⌋, \dots, ⌊ \frac{100^{2} + 100}{100} ⌋

where ⌊ x ⌋ is the greatest integer that is less than or equal to x

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

102 ?

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

there are totally 100 integers. so the number of distinct integers must be less than 100.

t h e r e a r e t o t a l l y 100 i n t e g e r s . s o t h e

n u m b e r o f d i s t i n c t i n t e g e r s m u s t b e

l e s s t h a n 100 .

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

You′re very right sir!I′ve committed mistake.

{Y o u}^{'} r e v e r y r i g h t s i r! I^{'} v e c o m m i t t e d

m i s t a k e .

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

⌊((1^2 +1)/(100))⌋, ⌊((2^2 +2)/(100))⌋, ⌊((3^2 +3)/(100))⌋, ..., ⌊((100^2 +100)/(100))⌋ ⌊((n^2 +n)/(100))⌋=⌊(( ((n(n+1))/2) )/((100)/2))⌋=⌊(( ((n(n+1))/2) )/(50))⌋ 0≤((n(n+1))/2)<50 ;n=1,2,...,9: ⌊(( ((n(n+1))/2) )/(50))⌋=0 50≤((n(n+1))/2)<100;n=10,11,...,13: ⌊(( ((n(n+1))/2) )/(50))⌋=1 .... ... 50k≤((n(n+1))/2)<50(k+1): ⌊(( ((n(n+1))/2) )/(50))⌋=k We′ve to solve many such inequalities... Continue...

⌊ \frac{1^{2} + 1}{100} ⌋, ⌊ \frac{2^{2} + 2}{100} ⌋, ⌊ \frac{3^{2} + 3}{100} ⌋, \dots, ⌊ \frac{100^{2} + 100}{100} ⌋

⌊ \frac{n^{2} + n}{100} ⌋ = ⌊ \frac{\frac{n (n + 1)}{2}}{\frac{100}{2}} ⌋ = ⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋

0 ⩽ \frac{n (n + 1)}{2} < 50; n = 1, 2, \dots, 9 :

⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋ = 0

50 ⩽ \frac{n (n + 1)}{2} < 100; n = 10, 11, \dots, 13 :

⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋ = 1

\dots .

\dots

50 k ⩽ \frac{n (n + 1)}{2} < 50 (k + 1) :

⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋ = k

{W e}^{'} v e t o s o l v e m a n y s u c h

i n e q u a l i t i e s \dots

C o n t i n u e \dots

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

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

≪ \underset{A N S W E R}{“ SOMETHING is better than {NOTHING}^{″}} ≫

⌊ \frac{1^{2} + 1}{100} ⌋, ⌊ \frac{2^{2} + 2}{100} ⌋, ⌊ \frac{3^{2} + 3}{100} ⌋, \dots, ⌊ \frac{100^{2} + 100}{100} ⌋

⌊ \frac{n^{2} + n}{100} ⌋ = ⌊ \frac{\frac{n (n + 1)}{2}}{\frac{100}{2}} ⌋ = ⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋

50 k ⩽ \frac{n (n + 1)}{2} < 50 (k + 1), k \in {0, 1, 2, \dots} :

⌊ \frac{\frac{n (n + 1)}{2}}{50} ⌋ = k

F o r \frac{N}{D \in N} t o b e i n t e g e r, N s h o u l d b e

i n t e g r a l m u l t i p l e o f D

\begin{array}{ccccccccccccccccccccccccccccccccccccccccccccccccccc} n & \underset{\overset{OR}{c u m u l a t i v e t o t a l}}{n (n + 1) / 2} \\ 1 & 1 \\ 2 & 3 \\ 3 & 6 \\ 4 & 10 \\ 5 & 15 \\ 6 & 21 \\ 7 & 28 \\ 8 & 36 \\ 9 & 45 < 50 (1) \\ 10 & 55 \\ 11 & 66 \\ 12 & 78 \\ 13 & 91 < 100 (2) \\ 14 & 105 \\ 15 & 120 \\ 16 & 136 < 150 (3) \\ 17 & 153 \\ 18 & 171 \\ 19 & 190 < 200 (4) \\ 20 & 210 \\ 21 & 231 < 250 (5) \\ 22 & 253 \\ 23 & 276 < 300 (6) \\ 24 & 300 \\ 25 & 325 < 350 (7) \\ 26 & 351 \\ 27 & 378 < 400 (8) \\ 28 & 406 \\ 29 & 435 < 450 (9) \\ 30 & 465 \\ 31 & 496 < 500 (10) \\ 32 & 528 < 550 (11) \\ 33 & 561 \\ 34 & 595 < 600 (12) \\ 35 & 630 < 650 (13) \\ 36 & 666 < 700 (14) \\ 37 & 703 \\ 38 & 741 < 750 (15) \\ 39 & 780 < 800 (16 \\ 40 & 820 < 850 (17) \\ 41 & 861 < 900 (18) \\ 42 & 903 \\ 43 & 946 < 950 (19) \\ 44 & 990 < 1000 (20) \\ 45 & 1035 < 1050 (21) \\ 46 & 1081 < 1100 (22) \\ 47 & 1128 < 1150 (23) \\ 48 & 1176 < 1200 (24) \\ 49 & 1225 < 1250 (25) \\ 50 & 1275 < 1300 (26) \end{array} \begin{array}{ccccccccccccccccccccccccccccccccccccccccccccccccccc} n & \underset{\overset{OR}{c u m u l a t i v e t o t a l}}{n (n + 1) / 2} \\ 51 & 1326 < 1350 (27) \\ 52 & 1378 < 1400 (28) \\ 53 & 1431 < 1450 (29) \\ 54 & 1485 < 1500 (30 \\ 55 & 1540 < 1550 (31) \\ 56 & 1596 < 1600 (32) \\ 57 & 1653 < 1700 (33) \\ 58 & 1711 < 1750 (34) \\ 59 & 1770 < 1800 (35) \\ 60 & 1830 < 1850 (36) \\ 61 & 1891 < 1900 (37) \\ 62 & 1953 < 2000 (38) \\ 63 & 2016 < 2050 (39) \\ 64 & 2080 < 2100 (40) \\ 65 & 2145 < 2150 (41) \\ 66 & 2211 < 2250 (42) \\ 67 & 2278 < 2300 (43) \\ 68 & 2346 < 2350 (44) \\ 69 & 2415 < 2450 (45) \\ 70 & 2485 < 2500 (46) \\ 71 & 2556 < 2600 (47) \\ 72 & 2628 < 2650 (48) \\ 73 & 2701 < 2750 (49) \\ 74 & 2775 < 2800 (50) \\ 75 & 2850 < 2900 (51) \\ 76 & 2926 < 2950 (52) \\ 77 & 3003 < 3050 (53) \\ 78 & 3081 < 3100 (54) \\ 79 & 3160 < 3200 (55) \\ 80 & 3240 < 3250 (56) \\ 81 & 3321 < 3350 (57) \\ 82 & 3403 < 3450 (58) \\ 83 & 3486 < 3500 (59) \\ 84 & 3570 < 3600 (60) \\ 85 & 3655 < 3700 (61) \\ 86 & 3741 < 3750 (62) \\ 87 & 3828 < 3850 (63) \\ 88 & 3916 < 3950 (64) \\ 89 & 4005 < 4050 (65) \\ 90 & 4095 < 4100 (66) \\ 91 & 4186 < 4200 (67) \\ 92 & 4278 < 4300 (68) \\ 93 & 4371 < 4400 (69) \\ 94 & 4465 < 4500 (70) \\ 95 & 4560 < 4600 (71) \\ 96 & 4656 < 4700 (72) \\ 97 & 4753 < 4800 (73) \\ 98 & 4851 < 4900 (74) \\ 99 & 4950 < 5000 (75) \\ 100 & \underset{\begin{matrix} 76 \end{matrix}}{5050 < 5100 (76)} \end{array}

C h a n g i n g c o l o u r o f f o n t h e r e

m e a n s^{'} o t h e r d i s t i n c t {i n t e g e r}^{'} .

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

sum (if (ceil (a (k)) < a (k + 1)), k = 0 . . 101)

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

Ans: 76 adhigenz sir, please confirm the answer before I give final shape to my solution.

A n s : 76

a d h i g e n z s i r, p l e a s e c o n f i r m t h e

a n s w e r b e f o r e I g i v e f i n a l s h a p e t o

m y s o l u t i o n .

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

a n s w e r 76 i s c o r r e c t s i r .

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

T h a n k s sir!

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

⌊((n^2 +n)/(100))⌋=k 1≤n≤100 ⇒0≤k≤101 but not each number from 0 to 101 is for k possible. k≤((n(n+1))/(100))<k+1 100k≤n(n+1)<100(k+1) n^2 +n−100k≥0 n≥((−1+(√(1+400k)))/2) n^2 +n−100(k+1)<0 n<((−1+(√(1+400(k+1))))/2) ((−1+(√(1+400k)))/2)≤n<((−1+(√(1+400(k+1))))/2) we can see a “k” is possible, i.e. a “n” exists, only if ⌈((−1+(√(1+400k)))/2)⌉<((−1+(√(1+400(k+1))))/2) for k from 0 to 101, only 76 numbers fulfill this condition. i checked this using Graph.

⌊ \frac{n^{2} + n}{100} ⌋ = k

1 ⩽ n ⩽ 100

\Rightarrow 0 ⩽ k ⩽ 101

b u t n o t e a c h n u m b e r f r o m 0 t o 101 i s

f o r k p o s s i b l e .

k ⩽ \frac{n (n + 1)}{100} < k + 1

100 k ⩽ n (n + 1) < 100 (k + 1)

n^{2} + n - 100 k ⩾ 0

n ⩾ \frac{- 1 + \sqrt{1 + 400 k}}{2}

n^{2} + n - 100 (k + 1) < 0

n < \frac{- 1 + \sqrt{1 + 400 (k + 1)}}{2}

\frac{- 1 + \sqrt{1 + 400 k}}{2} ⩽ n < \frac{- 1 + \sqrt{1 + 400 (k + 1)}}{2}

w e c a n s e e a “ k^{″} i s p o s s i b l e, i . e . a “ n^{″}

e x i s t s, o n l y i f

⌈ \frac{- 1 + \sqrt{1 + 400 k}}{2} ⌉ < \frac{- 1 + \sqrt{1 + 400 (k + 1)}}{2}

f o r k f r o m 0 t o 101, o n l y 76 n u m b e r s

f u l f i l l t h i s c o n d i t i o n . i c h e c k e d t h i s

u s i n g G r a p h .

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

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

you did a nice work sir using the table!

y o u d i d a n i c e w o r k s i r u s i n g t h e t a b l e!

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

T_{}^{H^{A} N} X

S_{I} R

Commented by Tawa11 last updated on 18/Sep/22

Great sirs

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